Lakshya Education MCQs

Question: Which of the following trigonometric expressions is equal to sec6 θ ?
Options:
A.tan6 θ+3 tan2 θ sec2 θ+1
B.tan6 θ−3 tan2 θ sec2 θ+1
C.tan6 θ−1
D.tan6 θ+1
Answer: Option A
: A

sec6θ=(sec2θ)3

=(tan2θ+1)3(1+tan2θ=sec2θ)

=tan6θ+3tan4θ+3tan2θ+1

=tan6θ+3tan2θ(tan2θ)+3tan2θ+1

=tan6θ+3tan2θ(sec2θ1)+3tan2θ+1

=tan6θ+3tan2θsec2θ3tan2θ+3tan2θ+1

=tan6θ+3tan2θsec2θ+1

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More Questions on This Topic :

Question 1. sin18cos72=

: sin18cos72
=sin(9072)cos72
=cos72cos72=1 as sin(90A)=cosA
Question 2. (1+tanθ+secθ)(1+cotθcosecθ)=

: C

(1+tanθ+secθ)(1+cotθcosecθ)
=(1+sinθcosθ+1cosθ)(1+cosθsinθ1sinθ)
=(cosθ+sinθ+1)cosθ×(sinθ+cosθ1)sinθ
=(cosθ+sinθ)212cosθsinθ
=(cos2θ+sin2θ+2cosθsinθ1)cosθsinθ
=(1+2cosθsinθ1)cosθsinθ
=2cosθsinθcosθsinθ=2
Question 3. If cos 3θ=32, 0° < 3θ < 90°, then find the value of θ.
  1.    15° 
  2.    10°
  3.    0°
  4.    12°
Answer: Option B
: B

Given: cos3θ=32
We know that cos30=32
Comparing the two we get,
3θ=30.... (given0 < 3θ < 90)
θ=10
Question 4. (1+tanA tanB)2+(tanA  tanB)2sec2A sec2B= ___
  1.    1
  2.    -tan A
  3.    2
  4.    cot A
Answer: Option A
: A

(1+tan A tan B)2+(tan A-tan B)2sec2Asec2B

=1+2tan A tan B+tan2Atan2B+tan2A2tan A tan B+tan2Bsec2Asec2B

=1+tan2Atan2B+tan2A+tan2Bsec2Asec2B

=1+tan2A+tan2B+tan2Atan2Bsec2Asec2B

=1(1+tan2A)+tan2B(1+tan2A)sec2Asec2B

=(1+tan2A)(1+tan2B)sec2Asec2B

=(sec2A)(sec2B)sec2Asec2B.(1+tan2θ=sec2θ)

=1
Question 5. If x=a cosec θ and y=b cot θ, then which of the following equations is true ?
  1.    x2−y2=a2−b2
  2.    x2a2−y2b2=1
  3.    x2+y2=a2+b2
  4.    x2a2+y2b2=1
Answer: Option B
: B

x=acosecθxa=cosecθ.....(1) y=bcotθyb=cotθ.....(2) Squaring the equations and subtracting (2) from (1) we get, x2a2y2b2=cosec2θcot2θ x2a2y2b2=1....(1+cot2θ=cosec2θ)

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