Question

2^5 (sinθ)^4 (cosθ)^2 =cos6θ -2cos4θ -cos2θ +2

Answer 1

Answer:

We have,

4cosθ+3sinθ=5

Then,

4cosθ+3sinθ=5

4cosθ+3

1−cos

2

θ

=5

3

1−cos

2

θ

=5−4cosθ

Squaring both side and we get,

9(1−cos

2

θ)=(5−4cosθ)

2

⇒9−9cos

2

θ=25+16cos

2

θ−40cosθ

⇒25+16cos

2

θ−40cosθ−9+9cos

2

θ=0

⇒25cos

2

θ−40cosθ+16=0

⇒(5cosθ)

2

−2×5cosθ×4+4

2

=0

⇒(5cosθ−4)

2

=0

⇒5cosθ−4=0

⇒5cosθ=4

⇒cosθ=

5

4

Put the value of given equation and we get,

4cosθ+3sinθ=5

⇒4×

5

4

+3sinθ=5

⇒

5

16

−5=3sinθ

⇒3sinθ=

5

16−25

⇒sinθ=−

15

9

⇒sinθ=

5

−3

Then,

tanθ=

cosθ

sinθ

tanθ=

5

4

−

5

3

tanθ=

4

−3

Hence, this is the answer.