Question

2+tan²theta + 1/ tan² theta = 1/sin²theta - sin^4 theta​

Answer 1

Answer:

Solution:

Given

LHS=\begin{gathered}\left(1+\frac{1}{tan^{2}\theta}\right)\left(1+\frac{1}{cot^{2}\theta}\right)\\ < /p > < p > = \left(1+cot^{2}\theta\right)\cdot\left(1+tan^{2}\theta\right)\end{gathered}

(1+

tan

2

θ

1

)(1+

cot

2

θ

1

)

</p><p>=(1+cot

2

θ)⋅(1+tan

2

θ)

_________________________

We know,

i) \frac{1}{tan\theta}=cot\theta

tanθ

1

=cotθ

ii)\frac{1}{cot\theta}=tan\theta

cotθ

1

=tanθ

Trigonometric identities :

1) 1+\cot^{2}\theta = cosec^{2}\theta1+cot

2

θ=cosec

2

θ

2) 1+tan^{2}\theta = sec^{2}\theta1+tan

2

θ=sec

2

θ

3) cos^{2}\theta=1-sin^{2}\thetacos

2

θ=1−sin

2

θ

and

a) cosec\theta = \frac{1}{sin\theta}cosecθ=

sinθ

1

b) sec\theta = \frac{1}{cos\theta}secθ=

cosθ

1

_________________________

=cosec^{2}\theta sec^{2}\thetacosec

2

θsec

2

θ

= \frac{1}{sin^{2}\theta}\cdot\frac{1}{cos^{2}\theta}

sin

2

θ

1

⋅

cos

2

θ

1

= \frac{1}{sin^{2}\theta\left(1-sin^{2}\theta\right)}

sin

2

θ(1−sin

2

θ)

1

= \frac{1}{sin^{2}\theta-sin^{4}\theta}

sin

2

θ−sin

4

θ

1

=RHSRHS

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