Practice Set 6.1 Trigonometry Class 10th Mathematics Part 2 MHB Solution

Practice Set 6.1

  1. If sintegrate heta = 7/25 find the values of cosθ and tanθ.
  2. sin^2theta /costheta +costheta = sectheta Prove that:
  3. If tantheta = 3/4 find the values of secθ and cosθ.
  4. If cottheta = 40/9 find the values of cosecθ and sinθ.
  5. If 5secθ- 12cosecθ = 0, find the values of secθ, cosθ and sinθ.
  6. If tanθ = 1 then, find the values of sintegrate heta +costheta /sectheta +cosectheta…
  7. cos^2theta (1+tan^2theta) = 1 Prove that:
  8. root 1-sintegrate heta /1+sintegrate heta = sectheta -tantheta Prove that:…
  9. (sec θ - cos θ) (cot θ + tan θ) = tan θ sec θ Prove that:
  10. cot θ + tan θ = cosec θ sec θ Prove that:
  11. 1/sectheta -tantheta = sectheta +tantheta Prove that:
  12. sin^4 θ - cos^4 θ = 1 - 2cos^2 θ Prove that:
  13. sectheta +tantheta = costheta /1-sintegrate heta Prove that:
  14. If tantheta + 1/tantheta = 2 then show that tan^2theta + 1/tan^2theta = 2 Prove that:…
  15. tana/(1+tan^2a)^2 + cota/(1+cot^2a)^2 = sinacosa Prove that:
  16. sec^4 A (1- sin^4 A) - 2tan^2 A = 1 Prove that:
  17. tantheta /sectheta -1 = tantheta +sectheta +1/tantheta +sectheta -1 Prove that:…

Practice Set 6.1
Question 1.

If  find the values of cosθ and tanθ.


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.

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