Practice Set 6.1

- If sintegrate heta = 7/25 find the values of cosθ and tanθ.
- sin^2theta /costheta +costheta = sectheta Prove that:
- If tantheta = 3/4 find the values of secθ and cosθ.
- If cottheta = 40/9 find the values of cosecθ and sinθ.
- If 5secθ- 12cosecθ = 0, find the values of secθ, cosθ and sinθ.
- If tanθ = 1 then, find the values of sintegrate heta +costheta /sectheta +cosectheta…
- cos^2theta (1+tan^2theta) = 1 Prove that:
- root 1-sintegrate heta /1+sintegrate heta = sectheta -tantheta Prove that:…
- (sec θ - cos θ) (cot θ + tan θ) = tan θ sec θ Prove that:
- cot θ + tan θ = cosec θ sec θ Prove that:
- 1/sectheta -tantheta = sectheta +tantheta Prove that:
- sin^4 θ - cos^4 θ = 1 - 2cos^2 θ Prove that:
- sectheta +tantheta = costheta /1-sintegrate heta Prove that:
- If tantheta + 1/tantheta = 2 then show that tan^2theta + 1/tan^2theta = 2 Prove that:…
- tana/(1+tan^2a)^2 + cota/(1+cot^2a)^2 = sinacosa Prove that:
- sec^4 A (1- sin^4 A) - 2tan^2 A = 1 Prove that:
- tantheta /sectheta -1 = tantheta +sectheta +1/tantheta +sectheta -1 Prove that:…

###### Practice Set 6.1

Answer:

We know that,

sin2θ + cos2θ = 1

Also,

Question 2.

Prove that:

Answer:

Taking LHS

[As, sin2θ + cos2θ = 1]

= sec θ

= RHS

Proved !

Question 3.

If find the values of secθ and cosθ.

Answer:

We know that,

sec2θ= 1 + tan2θ

Also,

Question 4.

If find the values of cosecθ and sinθ.

Answer:

We know that,

cosec2θ = 1 + cot2θ

⇒

Also,

∴ SinΘ=

Question 5.

If 5secθ- 12cosecθ = 0, find the values of secθ, cosθ and sinθ.

Answer:

5secθ - 12cosecθ = 0

⇒ 5secθ = 12cosecθ

As we have,

Also, We know that,

sec2θ= 1 + tan2θ

Also,

Now, again using

Question 6.

If tanθ = 1 then, find the values of

Answer:

Given,

tan θ = 1

⇒ θ = 45° [as tan 45° = 1]

Also,

= sinθ cosθ

= sin 45° cos 45°

Question 7.

Prove that:

Answer:

Taking LHS

cos2θ(1 + tan2θ)

= cos2θ sec2θ [As, sec2θ = 1 + tan2θ]

= 1

= RHS

Proved !

Question 8.

Prove that:

Answer:

Taking LHS

[(a + b)(a - b) = a2 - b2 ]

[As, sin2θ + cos2θ = 1]

= RHS

Proved !

Question 9.

Prove that:

(sec θ – cos θ) (cot θ + tan θ) = tan θ sec θ

Answer:

Taking LHS

(secθ - cosθ)(cotθ + tanθ)

= secθcotθ + secθtanθ - cosθcotθ - cosθtanθ

Taking LCM of first three terms,

[As, sin2θ + cos2θ = 1]

= tanθsecθ

= RHS

Proved !

Question 10.

Prove that:

cot θ + tan θ = cosec θ sec θ

Answer:

Taking LHS, and putting and

= cotθ + tanθ

[As, sin2θ + cos2θ = 1]

= RHS

Proved !

Question 11.

Prove that:

Answer:

Taking LHS

= secθ + tanθ [As, sec2θ = 1 + tan2θ ⇒ sec2θ - tan2θ = 1]

= RHS

Proved !

Question 12.

Prove that:

sin4 θ – cos4 θ = 1 – 2cos2 θ

Answer:

L.H.S = sin4θ – cos4θ

= (sin2θ – cos2θ)(sin2θ + cos2θ)

= (sin2θ – cos2θ)

= (1 – cos2θ – cos2θ)

= 1- 2cos2θ

Question 13.

Prove that:

Answer:

Taking RHS

(Multiplying both Numerator and Denominator by 1+SinΘ)

[As, sin2θ + cos2θ = 1]

= secθ + tanθ

= LHS

Proved !

Question 14.

Prove that:

If then show that

Answer:

Given,

Squaring both side,

[(a + b)2 = a2 + b2 + 2ab]

Hence, Proved !

Question 15.

Prove that:

Answer:

Taking RHS

= sinA cos3A + cosA sin3A

= sinAcosA(cos2A + sin2A) [As, sin2θ + cos2θ = 1]

= sinAcosA

= RHS

Proved !

Question 16.

Prove that:

sec4A (1– sin4A) – 2tan2 A = 1

Answer:

Taking LHS

= sec4A(1 - sin4A) - 2tan2A

= sec4A - sin4A sec4A - 2tan2A

= sec4A - tan4A - tan2A - tan2A

= sec4A - tan2A(1 + tan2A) - tan2A

= sec4A - tan2A sec2A - tan2A [As, sec2θ = 1 + tan2θ ]

= sec2A(sec2A - tan2A) - tan2A

= sec2A - tan2A [As, sec2θ = 1 + tan2θ ⇒ sec2θ - tan2θ = 1]

= 1

= RHS

Proved !

Question 17.

Prove that:

Answer:

Taking RHS

[As sec2θ -1 = tan2θ ]

= LHS

Proved.