Trigonometry Class 10th Mathematics AP Board Solution

Class 10th Mathematics AP Board Solution
Exercise 11.1
  1. In right angle triangle ABC, 8 cm, 15 cm and 17 cm are the lengths of AB, BC…
  2. The sides of a right angle triangle PQR are PQ = 7 cm, QR = 25 cm and ∠Q = 90°…
  3. In a right angle triangle ABC with right angle at B, in which a = 24 units, b =…
  4. If cosa = 12/13 then find sin A and tan A
  5. If 3 tanA = 4, then find sin A and cos A.
  6. If ∠A and ∠X are acute angles such that cos A = cos X then show that ∠A = ∠X.…
  7. Given cottheta = 7/8 then evaluate A. (1+sintegrate heta) (1-sintegrate…
  8. Given cottheta = 7/8 then evaluate B. (1+sintegrate heta)/costheta…
  9. In a right angle triangle ABC, right angle is at B, if tana = root 3 then find…
  10. In a right angle triangle ABC, right angle is at B, if tana = root 3 then find…
Exercise 11.2
  1. A. sin 45° + cos 45° Evaluate the following.
  2. B. cos45^circle /sec30^circle + cosec60^circle Evaluate the following.…
  3. C. sin30^circle +tan45^circle -cosec60^circle /cot45^circle +cos60^circle…
  4. D. 2tan^2 45° + cos^2 30° - sin^2 60° Evaluate the following.
  5. E. sin^260^circle - tan^260^circle /sin^230^circle + cos^230^circle Evaluate…
  6. A) 2tan30^circle /1+tan^245^circle A. sin 60° B. cos 60° C. tan 30° D. sin 30°…
  7. B) 1-tan^245^circle /1+tan^245^circle A. tan 90° B. 1 C. sin 45° D. 0 Choose…
  8. C) 2tan30^circle /1-tan^230^circle A. cos 60° B. sin 60° C. tan 60° D. sin 30°…
  9. Evaluate sin 60° cos 30° + sin 30°cos 60°. What is the value of sin(60° + 30°).…
  10. Is it right to say cos(60° + 30°) = cos 60° cos 30° - sin 60° sin 30°.…
  11. In right angle triangle ΔPQR, right angle is at Q and PQ = 6cms ∠RPQ = 60°.…
  12. In right angle is at Y, YZ = x and XZ = 2x then determine ∠YXZ and ∠YZX.…
  13. Is it right to say that sin(A + B) = Sin A + Sin B? Justify your answer.…
Exercise 11.3
  1. A) tan36^circle /cot54^circle Evaluate
  2. B) cos 12° - sin 78° Evaluate
  3. C) cosec 31° - sec 59° Evaluate
  4. D) sin 15° sec 75° Evaluate
  5. E. tan 26° tan 64° Evaluate
  6. A. tan 48° tan16° tan 42° tan 74° = 1 Show that
  7. B. cos36°cos54° - sin36°sin54° Show that
  8. If tan2A = cot(A - 18°) where 2A is an acute angle. Find the value of A.…
  9. If tan A = cot B where A and B are acute angles, prove that A + B = 90°…
  10. If A, B and C are interior angles of a triangle ABC, then show that tan (a+b/2)…
  11. Express sin 75° + cos 65° in terms of trigonometric ratios of angles between 0o…
Exercise 11.4
  1. A. (1 + tanθ + secθ)(1 + cotθ - cosecθ) Evaluate the following :
  2. B. (sinθ + cosθ)^2 + (sinθ - cosθ)^2 Evaluate the following :
  3. C. (sec^2 θ - 1) (cosec^2 θ - 1) Evaluate the following :
  4. Show that (cosectheta +cottheta)^2 = 1-costheta /1+costheta
  5. Show that root 1+sina/1-sina = seca+tana
  6. Show that 1-tan^2a/cot^2a-1 = tan^2a
  7. Show that 1/costheta -costheta = tantheta sintegrate heta
  8. Simplify secA(1 - sinA)(secA + tanA)
  9. Prove that (sin A + cosec A)^2 + (cos A + sec A)2 = 7 + tan^2 A + cot^2 A…
  10. Simplify (1 - cosθ) (1 + cosθ) (1 + cot^2 θ)
  11. If secθ + tanθ = p, then what is the value of secθ - tanθ ?
  12. If cosesθ + cotθ = k then prove that costheta = k^2 - 1/k^2 + 1

Exercise 11.1
Question 1.

In right angle triangle ABC, 8 cm, 15 cm and 17 cm are the lengths of AB, BC and CA respectively. Then, find out sin A, cos A and tan A.


Answer:

We have



Given: ∠ABC = 90°, AB = 8 cm, BC = 15 cm and CA = 17 cm


We know sin A is given by,



⇒  [∵, perpendicular is the side opposite to the angle A & hypotenuse is the side opposite to the right angle of that triangle]


⇒  …(i)


Also, cos A is given by



⇒ 


⇒  …(ii)


Now, tan A can be found out by two ways:


Method 1: tan A is given by,



⇒ 


⇒ 


Method 2: tan A can also be written as,



⇒  [from equations (i) & (ii)]


⇒ 


Thus,  and .



Question 2.

The sides of a right angle triangle PQR are PQ = 7 cm, QR = 25 cm and ∠Q = 90° respectively. Then find, tan P – tan R.


Answer:

We have


We don’t need to find hypotenuse (PR) in the ∆PQR as tan θ = perpendicular/base.



⇒ 


And 


⇒ 


tan P – tan R = 


⇒ tan P – tan R = 


⇒ tan P – tan R = 576/175


Thus, tan P – tan R = 576/175



Question 3.

In a right angle triangle ABC with right angle at B, in which a = 24 units, b = 25 units and ∠BAC = θ. Then, find cosθ and tanθ.


Answer:

We have


In ∆ABC, ∠ABC = 90° and ∠BAC = θ.


Using this information, we can say


AC = hypotenuse of the triangle


BC = perpendicular (side opposite to the angle θ or ∠BAC)


Using Pythagoras theorem,


(hypotenuse)2 = (perpendicular)2 + (base)2


⇒ (25)2 = (24)2 + (base)2


⇒ (AB)2 = 625 – 576


⇒ (AB)2 = 49


⇒ AB = √49 = 7 units


So, we have AB = 7 units, BC = 24 units and AC = 25 units.


Thus, 


⇒ 


And 


⇒ 


Thus,  and .



Question 4.

If  then find sin A and tan A


Answer:

Given that,


But 


⇒ 


⇒ base = 12 and hypotenuse = 13


So, using Pythagoras theorem, we can say


(hypotenuse)2 = (perpendicular)2 + (base)2


⇒ (perpendicular)2 = (hypotenuse)2 – (base)2


⇒ (perpendicular)2 = (13)2 – (12)2


⇒ (perpendicular)2 = 169 – 144 = 25


⇒ perpendicular = √25 = 5


Using perpendicular = 5, base = 12 and hypotenuse = 13, we can find out sin A and tan A.


Sin A is given by



⇒ 


And, tan A is given by



⇒ 


Thus,  and .



Question 5.

If 3 tanA = 4, then find sin A and cos A.


Answer:

Given that, 3 tan A = 4

⇒ 


But 


⇒ 


⇒ perpendicular = 4 and base = 3


So, using Pythagoras theorem, we can say


(hypotenuse)2 = (perpendicular)2 + (base)2


⇒ (hypotenuse)2 = (4)2 + (3)2


⇒ (hypotenuse)2 = 16 + 9 = 25


⇒ hypotenuse = √25 = 5


Using perpendicular = 4, base = 3 and hypotenuse = 5, we can find out sin A and cos A.


Sin A is given by



⇒ 


And, cos A is given by



⇒ 


Thus,  and .



Question 6.

If ∠A and ∠X are acute angles such that cos A = cos X then show that ∠A = ∠X.


Answer:

We have


In this ∆AOX,


cos A = cos X


and cos A is given by,


 [∵, ]


Similarly, 


⇒ 


⇒ AO = OX [clearly, since denominator from either sides cancel each other]


Now, if sides AO and OX of ∆AOX are equal.


Then, ∠A = ∠X [∵, In a triangle, angles opposite to the equal sides are also equal]


Hence, ∠A = ∠X



Question 7.

Given  then evaluate

A. 


Answer:

We have been given that,



And we have to solve for .


We know the formula: (x + y)(x – y) = x2 – y2


Using this,



Also, we know the relationship between cos θ and sin θ which is given by


cos2 θ + sin2 θ = 1


⇒ cos2 θ = 1 – sin2 θ


So,


⇒  

As we know,



So, B = 7 and P = 8

By Pythagoras theorem,

H2 = P2 + B2
H2 = 82 + 72
= 64 + 49
=113

H = √113

As we know,



So,



Now,








 
Question 8.

Given  then evaluate

B. 


Answer:

Given, 

We know that, cosec2θ = 1 + cot2θ


Now, To Find:

Dividing both numerator and denominator by sin θ, we have
 
Question 9.

In a right angle triangle ABC, right angle is at B, if  then find the value of

A. sin A cosC + cos A sin C


Answer:

We have



Given that, tan A = √3/1


And tan A is given by,



⇒ 


⇒ perpendicular = √3x and base = x


Then, we can use Pythagoras theorem in ∆ABC,


(hypotenuse)2 = (perpendicular)2 + (base)2


⇒ (hypotenuse)2 = (√3x)2 + (x)2


⇒ (hypotenuse)2 = 3x2 + x2 = 4x2


⇒ hypotenuse = √(4x2) = 2x


We have, AB = √3x, BC = x and AC = 2x.


Using these values,



⇒ 


⇒ 


⇒  …(i)


Similarly, 


⇒ 


⇒  …(ii)


Also,



⇒ 


⇒ 


⇒  …(iii)


Similarly, 


⇒ 


⇒  …(iv)


We have to solve: sin A cos C + cos A sin C.


Substituting equations (i), (ii), (iii) & (iv) in above,


sin A cos C + cos A sin C = 


⇒ sin A cos C + cos A sin C = 1/4 + 3/4


⇒ sin A cos C + cos A sin C = 4/4 = 1


Thus, sin A cos C + cos A sin C = 1.



Question 10.

In a right angle triangle ABC, right angle is at B, if  then find the value of

B. cos A cos C – sin A sin C


Answer:

To find: cos A cos C – sin A sin C.


From previous part of the question, we have






Using these values, we get


cos A cos C – sin A sin C = 


⇒ cos A cos C – sin A sin C = √3/4 - √3/4 = 0


Thus, cos A cos C – sin A sin C = 0.




Exercise 11.2
Question 1.

Evaluate the following.

A. sin 45° + cos 45°


Answer:

By trigonometric identities, we can say


sin 45° = 1/√2


and cos 45° = 1/√2


Adding them, we get


sin 45° + cos 45° = 1/√2 + 1/√2


⇒ sin 45° + cos 45° = 2/√2 = √2


Thus, sin 45° + cos 45° = √2.



Question 2.

Evaluate the following.

B. 


Answer:



Trigonometric identities:

cos 45° = 1/√2

sec 30° = 1/cos 30° = 2/√3 [∵, cos 30° = √3/2]

cosec 60° = 1/sin 60° = 2/√3 [∵, sin 60° = √3/2]

Putting the values we get,


Question 3.

Evaluate the following.

C. 


Answer:

By trigonometric identities,


cot 45° = 1/tan 45° = 1/1 = 1


sec 30° = 1/cos 30° = 2/√3 [∵, cos 30° = √3/2]


cosec 60° = 1/sin 60° = 2/√3 [∵, sin 60° = √3/2]


sin 30° = 1/2


tan 45° = 1


cos 60° = 1/2


Putting all these values in , we get



Since, numerator is equal to denominator in the above calculation, we can say



Question 4.

Evaluate the following.

D. 2tan2 45° + cos2 30° - sin2 60°


Answer:

By trigonometric identities,

sin 60° = √3/2


tan 45° = 1


cos 30° = √3/2


Putting these values in 2 tan2 45° + cos2 30° - sin2 60°, we get


2 tan2 45° + cos2 30° - sin2 60° = 2(1)2 + (√3/2)2 – (√3/2)2


⇒ 2 tan2 45° + cos2 30° - sin2 60° = 2


Thus, 2 tan2 45° + cos2 30° - sin2 60° = 2.



Question 5.

Evaluate the following.

E. 


Answer:

We have to solve:



Recall the trigonometric identities,


sin2 θ + cos2 θ = 1


& sec2 α – tan2 α = 1


Put θ = 30° and α = 60°, we get


sin2 30° + cos2 30° = 1


& sec2 60° - tan2 60° = 1


So,



Thus,



Question 6.

Choose the right option and justify your choice -

A) 

A. sin 60°

B. cos 60°

C. tan 30°

D. sin 30°


Answer:

we know,

tan 30° = 1/√3


tan 45° = 1


Then, putting these values in the question, we get



⇒ 


And we know, tan 30° = 1/√3


But, sin 60° = √3/2


cos 60° = 1/2


& sin 30° = 1/2


Thus, option (C) is correct.


Question 7.

Choose the right option and justify your choice -

B) 

A. tan 90°

B. 1

C. sin 45°

D. 0


Answer:

We know that,

tan 45° = 1


So, using this value of tangent, we can write



The answer has come out to be 0.


Thus, option (D) is correct.


Question 8.

Choose the right option and justify your choice -

C) 

A. cos 60°

B. sin 60°

C. tan 60°

D. sin 30°


Answer:

We know that,

tan 30° = 1/√3


So, using this value of tangent, we can write



⇒ 


⇒ 


⇒ 


⇒ 


⇒ 


And we know, tan 60° = √3


But, cos 60° = 1/2


sin 60° = √3/2


& sin 30° = 1/2


Thus, option (C) is correct.


Question 9.

Evaluate sin 60° cos 30° + sin 30°cos 60°. What is the value of sin(60° + 30°). What can you conclude?


Answer:

Let us first solve sin 60° cos 30° + sin 30° cos 60°.

We know,


sin 60° = √3/2


cos 30° = √3/2


sin 30° = 1/2


& cos 60° = 1/2


So, sin 60° cos 30° + sin 30° cos 60° = 


⇒ sin 60° cos 30° + sin 30° cos 60° = 


⇒ sin 60° cos 30° + sin 30° cos 60° = 1 …(i)


Now, for sin (60° + 30°):


sin (60° + 30°) = sin 90°


⇒ sin (60° + 30°) = 1 [∵, sin 90° = 1] …(ii)


By equations (i) & (ii), we can conclude that


sin (60° + 30°) = sin 60° cos 30° + sin 30° cos 60°


And infact in general, let 60° = x and 30° = y. Then,


sin (x + y) = sin x cos y + sin y cos x



Question 10.

Is it right to say cos(60° + 30°) = cos 60° cos 30° - sin 60° sin 30°.


Answer:

Let us solve Left-Hand-Side:

cos (60° + 30°) = cos 90°


⇒ cos (60° + 30°) = 0 [∵, cos 90° = 0]


Now, solve for Right-Hand-Side:



⇒ 


⇒ cos 60° cos 30° - sin 60° sin 30° = 0



Question 11.

In right angle triangle ΔPQR, right angle is at Q and PQ = 6cms ∠RPQ = 60°. Determine the lengths of QR and PR.


Answer:


To find QR:


Since, tan θ = perpendicular/base


We know that,



⇒  [∵, PQ = 6 cm & tan 60° = √3]


⇒ QR = 6√3


Now, PR can be found by two ways -


1st method: In ∆PQR, using Pythagoras theorem,


PR2 = PQ2 + QR2 [∵, (hypotenuse)2 = (perpendicular)2 + (base)2]


⇒ PR2 = 62 + (6√3)2


⇒ PR2 = 36 + 108 = 144


⇒ PR = √144 = 12


2nd Method:


Since, cos θ = base/hypotenuse


We know that,



⇒ 


⇒ 


⇒ PR = 2 × PQ


⇒ PR = 2 × 6 = 12


Thus, QR = 6√3 cm and PR = 12 cm.



Question 12.

In right angle is at Y, YZ = x and XZ = 2x then determine ∠YXZ and ∠YZX.


Answer:

We have


With given values, YZ = x and XZ = 2x, we can find out both angles.


For ∠YXZ:


Let ∠YXZ = θ, then



⇒ 


⇒ 


⇒ θ = sin-1(1/2)


⇒ θ = 30° [∵, sin 30° = 1/2]


For ∠YZX = α, then



⇒ 


⇒ 


⇒ α = cos-1(1/2)


⇒ α = 60° [∵, cos 60° = 1/2]



Question 13.

Is it right to say that sin(A + B) = Sin A + Sin B? Justify your answer.


Answer:

No, it is not correct to say that sin (A + B) = sin A + sin B.

Justification: Let’s justify it by showing contradiction.


Let it be true that, sin (A + B) = sin A + sin B.


Now, let A = 30° and B = 60°


Then,


sin (30° + 60°) = sin 30° + sin 60°


⇒ sin 90° = sin 30° + sin 60°


⇒ 1 = 1/2 + √3/2


⇒ 1 = (1 + √3)/2


But, it’s not true.


1 ≠ (1 + √3)/2


Hence, we have a contradiction.


And therefore, it’s not right to say that sin (A + B) = sin A + sin B.




Exercise 11.3
Question 1.

Evaluate

A) 


Answer:

By trigonometric identity, we have


tan (90° - θ) = cot θ


Replace θ = 54°


⇒ tan (90° - 54°) = cot 54°


⇒ tan 36° = cot 54° [∵, 90° - 54° = 36°]


Now, 


⇒ 



Question 2.

Evaluate

B) cos 12° - sin 78°


Answer:

By trigonometric identity, we can say

cos (90° - θ) = sin θ


Now, just replace θ by 78°.


We get


cos (90° - 78°) = sin 78°


⇒ cos 12° = sin 78° [∵, 90° - 78° = 12°]


Now, cos 12° - sin 78° = sin 78° - sin 78°


⇒ cos 12° - sin 78° = 0



Question 3.

Evaluate

C) cosec 31° - sec 59°


Answer:

By trigonometric identity, we can say

cosec (90° - θ) = sec θ


Replace θ by 59°.


We get


cosec (90° - 59°) = sec 59°


⇒ cosec 31° = sec 59°


Now, cosec 31° - sec 59° = sec 59° - sec 59°


⇒ cosec 31° - sec 59° = 0



Question 4.

Evaluate

D) sin 15° sec 75°


Answer:

By trigonometric identities, we can say

sin (90° - θ) = cos θ


And sec θ = 1/cos θ


Replace θ by 75°.


We get


sin (90° - 75°) = cos 75°


⇒ sin 15° = cos 75° [∵, 90° - 75° = 15°]


And sec 75° = 1/cos 75°


Using these values, we can solve the given expression.


sin 15° sec 75° = sin 15°/cos 75°


⇒ sin 15° sec 75° = cos 75°/cos 75°


⇒ sin 15° sec 75° = 1



Question 5.

Evaluate

E. tan 26° tan 64°


Answer:

By trigonometric identities, we can say


tan (90° - θ) = cot θ


And tan θ = 1/cot θ


Replace θ by 64°.


We get


tan (90° - 64°) = cot 64°


⇒ tan 26° = cot 64° [∵, 90° - 64° = 26°]


And tan 64° = 1/cot 64°


Using these values, we can solve for the given expression.


tan 26° tan 64° = cot 64° tan 64°


⇒ tan 26° tan 64° = cot 64°/cot 64°


⇒ tan 26° tan 64° = 1



Question 6.

Show that

A. tan 48° tan16° tan 42° tan 74° = 1


Answer:

We have


LHS = tan 48° tan 16° tan 42° tan 74° = (tan 48° tan 42°)(tan 16° tan 74°)


We know the trigonometric identities, we can say


tan (90° - θ) = cot θ


And tan θ = 1/cot θ


First, replace θ by 42°.


tan (90° - 42°) = cot 42°


⇒ tan 48° = cot 42° …(1)


And tan 42° = 1/cot 42° …(2)


Now, replace θ by 74°.


tan (90° - 74°) = cot 74°


⇒ tan 16° = cot 74° …(3)


And tan 74° = 1/cot 74° …(4)


Using equations (1), (2), (3) & (4), we get


LHS = tan 48° tan 16° tan 42° tan 74° = (tan 48° tan 42°)(tan 16° tan 74°)


= (cot 42°/cot 42°)(cot 74°/cot 74°)


= 1 = RHS


Thus, tan 48° tan 16° tan 42° tan 74° = 1.



Question 7.

Show that

B. cos36°cos54° - sin36°sin54°


Answer:

We have

LHS = cos 36° cos 54° - sin 36° sin 54°


We know the trigonometric identity,


cos (90° - θ) = sin θ


First, replace θ by 54°.


We get, cos (90° - 54°) = sin 54°


⇒ cos 36° = sin 54° …(1)


Now, replace θ by 36°.


We get, cos (90° - 36°) = sin 36°


⇒ cos 54° = sin 36° …(2)


Using equations (1) & (2), we get


LHS = cos 36° cos 54° - sin 36° sin 54° = sin 54° sin 36° - sin 54° sin 36°


= 0 = RHS


Thus, cos 36° cos 54° - sin 36° sin 54° = 0.



Question 8.

If tan2A = cot(A – 18°) where 2A is an acute angle. Find the value of A.


Answer:

Given that, 2A is an acute angle.

⇒ 2A < 90°


So, using trigonometric identity, we can say that


cot (90° - 2A) = tan 2A [∵, cot (90° - θ) = tan θ]


Now, replace tan 2A by cot (90° - 2A) in the given question.


tan 2A = cot (A – 18°)


⇒ cot (90° - 2A) = cot (A – 18°)


Now, we can compare the degrees from above, we get


90° - 2A = A – 18°


⇒ 2A + A = 90° + 18°


⇒ 3A = 108°


⇒ A = 108°/3


⇒ A = 36°


Thus, the value of A is 36°.



Question 9.

If tan A = cot B where A and B are acute angles, prove that A + B = 90°


Answer:

Given that, A and B are acute angles.

⇒ A < 90° & B < 90°


So, using trigonometric identity, we can say


tan (90° - B) = cot B [∵, tan (90° - θ) = cot θ]


Replace cot B of RHS by tan (90° - B) in the given question.


tan A = cot B


⇒ tan A = tan (90° - B)


Now, comparing the degrees from the above, we get


A = 90° - B


⇒ A + B = 90°


Hence, proved that A + B = 90°.



Question 10.

If A, B and C are interior angles of a triangle ABC, then show that 


Answer:

If A, B and C are interior angles of ∆ABC, then we can say that

A + B + C = 180° (by angle sum property of a triangle)


⇒ A + B = 180° - C


Take LHS:



⇒ 


⇒ 


[∵, tan (90° - θ) = cot θ, where θ = C/2 here]


Hence, .



Question 11.

Express sin 75° + cos 65° in terms of trigonometric ratios of angles between 0o and 45o.


Answer:

Given: sin 75° + cos 65°.

We can write,


75° = 90° - 15°


& 65° = 90° - 25°


Then, sin 75° + cos 65° = sin (90° - 15°) + cos (90° - 25°)


⇒ sin 75° + cos 65° = cos 15° + sin 25°


[∵, sin (90° - θ ) = cos θ & cos (90° - θ) = sin θ]


In cos 15° + sin 25°, 15° & 25° both are angles between 0° and 45°.


Thus, answer is sin 75° + cos 65° = cos 15° + sin 25°.




Exercise 11.4
Question 1.

Evaluate the following :

A. (1 + tanθ + secθ)(1 + cotθ – cosecθ)


Answer:

We have



⇒ 


⇒ 


⇒  [∵, (a + b)(a – b) = a2 – b2]


⇒  [∵, (a + b)2 = a2 + b2 + 2ab]


⇒  [∵, sin2 θ + cos2 θ = 1]


⇒ 


Thus, (1 + tan θ + sec θ)(1 + cot θ – cosec θ) = 2.



Question 2.

Evaluate the following :

B. (sinθ + cosθ)2 + (sinθ - cosθ)2


Answer:

We have

(sin θ + cos θ)2 + (sin θ – cos θ)2 = ((sin θ + cos θ) + (sin θ – cos θ))2 – 2(sin θ + cos θ)(sin θ – cos θ) [∵, a2 + b2 = (a + b)2 – 2ab]


⇒ (sin θ + cos θ)2 + (sin θ – cos θ)2 = (sin θ + cos θ + sin θ – cos θ)2 – 2(sin2 θ – cos2 θ)


[∵, (a + b)(a – b) = a2 – b2]


⇒ (sin θ + cos θ)2 + (sin θ – cos θ)2 = (2 sin θ)2 – 2 sin2 θ + 2 cos2 θ


⇒ (sin θ + cos θ)2 + (sin θ – cos θ)2 = 4 sin2 θ – 2 sin2 θ + 2 cos2 θ


⇒ (sin θ + cos θ)2 + (sin θ – cos θ)2 = 2 sin2 θ + 2 cos2θ


⇒ (sin θ + cos θ)2 + (sin θ – cos θ)2 = 2 (sin2 θ + cos2 θ)


⇒ (sin θ + cos θ)2 + (sin θ – cos θ)2 = 2 [∵, sin2 θ + cos2 θ = 1]


Thus, (sin θ + cos θ)2 + (sin θ – cos θ)2 = 2.



Question 3.

Evaluate the following :

C. (sec2θ – 1) (cosec2θ – 1)


Answer:

We have


⇒ 


⇒  [∵, (1 – cos2 θ) = sin2 θ & (1 – sin2 θ = cos2 θ]


⇒ (sec2 θ – 1)(cosec2 θ – 1) = 1


Thus, (sec2 θ – 1)(cosec2 θ – 1) = 1.



Question 4.

Show that 


Answer:

Using trigonometric identities,

cosec θ = 1/sin θ & cot θ = cos θ/sin θ


LHS = (cosec θ – cot θ)2





As we know, sin2 θ = 1 – cos2 θ


And (1 – cos2 θ) = (1 + cos θ)(1 – cos θ) [by (a2 – b2) = (a + b)(a – b)]


⇒ sin2 θ = (1 + cos θ)(1 – cos θ) …(i)


So, using equation (i),


LHS = (cosec θ – cot θ)2 = 


 = RHS


Hence, we have got


(cosec θ – cot θ)2 = .



Question 5.

Show that 


Answer:

Using trigonometric identity,


sin2 A + cos2 A = 1


⇒ cos2 A = 1 – sin2 A


Take Left hand side:


LHS = 






= sec A + tan A = RHS


Thus, .



Question 6.

Show that 


Answer:

Using trigonometric identity, we have

cot A = 1/tan A


Take left hand side,


LHS = 





= tan2 A = RHS


Thus,  = tan2 A.



Question 7.

Show that 


Answer:

Take left hand side of the given question:

LHS = 1/cos θ – cos θ


= (1 – cos2 θ)/cos θ


= sin2 θ/cos θ [∵, sin2 θ + cos2 θ = 1 ⇒ sin2 θ = 1 – cos2 θ]


= sin θ × sin θ/cos θ


= sin θ × tan θ [∵, tan θ = sin θ/cos θ]


= tan θ sin θ = RHS


Thus, .



Question 8.

Simplify secA(1 – sinA)(secA + tanA)


Answer:

By trigonometric identities, sec A = 1/cos A & tan A = sin A/cos A

Using these identities, we have


sec A (1 – sin A)(sec A + tan A) = 


⇒ sec A (1 – sin A)(sec A + tan A) = 


⇒ sec A (1 – sin A)(sec A + tan A) = 


⇒ sec A (1 – sin A)(sec A + tan A) = 


⇒ sec A (1 – sin A)(sec A + tan A) =  [∵, sin2 A + cos2 A = 1 ⇒ cos2 A = 1 – sin2 A]


⇒ sec A (1 – sin A)(sec A + tan A) = 1


Thus, sec A (1 – sin A)(sec A + tan A) = 1.



Question 9.

Prove that (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A


Answer:

Take left hand side of the given equation:

LHS = (sin A + cosec A)2 + (cos A + sec A)2

Expanding the squares by formula: (a + b)2 = a2 + b2 + 2ab

= sin2 A + cosec2 A + 2 sin A cosec A + cos2 A + sec2 A + 2 cos A sec A

Rearranging the terms, we get,

= (sin2 A + cos2 A) + 2 sin A cosec A + 2 cos A sec A + cosec2 A + sec2 A

we know that,




= 5 + 1/(sin2 A cos2 A) …(i)


Now, take right hand side of the equation:


RHS = 7 + tan2 A + cot2 A






= 5 + 1/(sin2 A cos2 A) …(ii)


From equation (i) & (ii),


LHS = RHS


Hence, proved.


Question 10.

Simplify (1 – cosθ) (1 + cosθ) (1 + cot2θ)


Answer:

We have

(1 – cos θ)(1 + cos θ)(1 + cot2 θ) = [(1 – cos θ)(1 + cos θ)](1 + cot2 θ)


⇒ (1 – cos θ)(1 + cos θ)(1 + cot2 θ) = (1 – cos2 θ)(1 + cot2 θ) [∵, (a + b)(a – b) = a2 – b2]


⇒ (1 – cos θ)(1 + cos θ)(1 + cot2 θ) = sin2 θ × (1 + cos2 θ/sin2 θ) [∵, (1 – cos2 θ) = sin2 θ & cot2 θ = cos2 θ/sin2 θ]


⇒ (1 – cos θ)(1 + cos θ)(1 + cot2 θ) = sin2 θ × (sin2 θ + cos2 θ)/sin2 θ


⇒ (1 – cos θ)(1 + cos θ)(1 + cot2 θ) = sin2 θ + cos2 θ


⇒ (1 – cos θ)(1 + cos θ)(1 + cot2 θ) = 1 [∵, sin2 θ + cos2 θ = 1]


Thus, (1 – cos θ)(1 + cos θ)(1 + cot2 θ) = 1.



Question 11.

If secθ + tanθ = p, then what is the value of secθ – tanθ ?


Answer:

Given that, sec θ + tan θ = p.

By trigonometric identity, we have


sec2 θ – tan2 θ = 1


So, sec2 θ – tan2 θ = 1


⇒ (sec θ – tan θ) (sec θ + tan θ) = 1


⇒ sec θ – tan θ = 1/(sec θ + tan θ)


⇒ sec θ – tan θ = 1/p [given]


Hence, sec θ – tan θ = 1/p.



Question 12.

If cosesθ + cotθ = k then prove that 


Answer:

Given that, cosec θ + cot θ = k

⇒  [∵, cosec θ = 1/sin θ & cot θ = cos θ/sin θ]


⇒ (1 + cos θ)/sin θ = k


⇒ 1 + cos θ = k sin θ


Squaring both sides, we get


(1 + cos θ)2 = (k sin θ)2


⇒ (1 + cos θ)2 = k2 sin2 θ


⇒ (1 + cos θ)2 = k2 (1 – cos2 θ) [∵, sin2 θ + cos2 θ = 1 ⇒ sin2 θ = 1 – cos2 θ]


⇒ (1 + cos θ)2 = k2 (1 – cos θ) (1 + cos θ) [∵, a2 – b2 = (a + b) (a – b)]


⇒ 1 + cos θ = k2 (1 – cos θ)


⇒ 1 + cos θ = k2 – k2 cos θ


⇒ k2 cos θ + cos θ = k2 – 1


⇒ cos θ (k2 + 1) = k2 – 1


⇒ cos θ = (k2 – 1)/(k2 + 1)


Thus, cos θ = (k2 – 1)/(k2 + 1).


Hence, proved.


PDF FILE TO YOUR EMAIL IMMEDIATELY PURCHASE NOTES & PAPER SOLUTION. @ Rs. 50/- each (GST extra)

HINDI ENTIRE PAPER SOLUTION

MARATHI PAPER SOLUTION

SSC MATHS I PAPER SOLUTION

SSC MATHS II PAPER SOLUTION

SSC SCIENCE I PAPER SOLUTION

SSC SCIENCE II PAPER SOLUTION

SSC ENGLISH PAPER SOLUTION

SSC & HSC ENGLISH WRITING SKILL

HSC ACCOUNTS NOTES

HSC OCM NOTES

HSC ECONOMICS NOTES

HSC SECRETARIAL PRACTICE NOTES

2019 Board Paper Solution

HSC ENGLISH SET A 2019 21st February, 2019

HSC ENGLISH SET B 2019 21st February, 2019

HSC ENGLISH SET C 2019 21st February, 2019

HSC ENGLISH SET D 2019 21st February, 2019

SECRETARIAL PRACTICE (S.P) 2019 25th February, 2019

HSC XII PHYSICS 2019 25th February, 2019

CHEMISTRY XII HSC SOLUTION 27th, February, 2019

OCM PAPER SOLUTION 2019 27th, February, 2019

HSC MATHS PAPER SOLUTION COMMERCE, 2nd March, 2019

HSC MATHS PAPER SOLUTION SCIENCE 2nd, March, 2019

SSC ENGLISH STD 10 5TH MARCH, 2019.

HSC XII ACCOUNTS 2019 6th March, 2019

HSC XII BIOLOGY 2019 6TH March, 2019

HSC XII ECONOMICS 9Th March 2019

SSC Maths I March 2019 Solution 10th Standard11th, March, 2019

SSC MATHS II MARCH 2019 SOLUTION 10TH STD.13th March, 2019

SSC SCIENCE I MARCH 2019 SOLUTION 10TH STD. 15th March, 2019.


SSC SCIENCE II MARCH 2019 SOLUTION 10TH STD. 18th March, 2019.

SSC SOCIAL SCIENCE I MARCH 2019 SOLUTION20th March, 2019

SSC SOCIAL SCIENCE II MARCH 2019 SOLUTION, 22nd March, 2019

XII CBSE - BOARD - MARCH - 2019 ENGLISH - QP + SOLUTIONS, 2nd March, 2019

HSC Maharashtra Board Papers 2020

(Std 12th English Medium)

HSC ECONOMICS MARCH 2020

HSC OCM MARCH 2020

HSC ACCOUNTS MARCH 2020

HSC S.P. MARCH 2020

HSC ENGLISH MARCH 2020

HSC HINDI MARCH 2020

HSC MARATHI MARCH 2020

HSC MATHS MARCH 2020

SSC Maharashtra Board Papers 2020

(Std 10th English Medium)

English MARCH 2020

HindI MARCH 2020

Hindi (Composite) MARCH 2020

Marathi MARCH 2020

Mathematics (Paper 1) MARCH 2020

Mathematics (Paper 2) MARCH 2020

Sanskrit MARCH 2020

Sanskrit (Composite) MARCH 2020

Science (Paper 1) MARCH 2020

Science (Paper 2)

Geography Model Set 1 2020-2021

MUST REMEMBER THINGS on the day of Exam

Are you prepared? for English Grammar in Board Exam.

Paper Presentation In Board Exam

How to Score Good Marks in SSC Board Exams

Tips To Score More Than 90% Marks In 12th Board Exam

How to write English exams?

How to prepare for board exam when less time is left

How to memorise what you learn for board exam

No. 1 Simple Hack, you can try out, in preparing for Board Exam

How to Study for CBSE Class 10 Board Exams Subject Wise Tips?

JEE Main 2020 Registration Process – Exam Pattern & Important Dates


NEET UG 2020 Registration Process Exam Pattern & Important Dates

How can One Prepare for two Competitive Exams at the same time?

8 Proven Tips to Handle Anxiety before Exams!

BUY FROM PLAY STORE

DOWNLOAD OUR APP

HOW TO PURCHASE OUR NOTES?

S.P. Important Questions For Board Exam 2021

O.C.M. Important Questions for Board Exam. 2021

Economics Important Questions for Board Exam 2021

Chemistry Important Question Bank for board exam 2021

Physics – Section I- Important Question Bank for Maharashtra Board HSC Examination

Physics – Section II – Science- Important Question Bank for Maharashtra Board HSC 2021 Examination