Question

Prove that tan²θ−sin²θ= sin⁴θsec²θ​

Answer 1

Answer:

here is ur answer

hope this helps you

Answer 2

ANSWER:

To Prove:

• tan²θ − sin²θ = sin⁴θsec²θ

Solution:

We need to prove that,

$\implies\tan^2\theta-\sin^2\theta=\sin^4\theta\sec^2\theta$

Taking LHS,

$\implies\tan^2\theta-\sin^2\theta$

We know that,

$\hookrightarrow\tan A=\dfrac{\sin A}{\cos A}$

So,

$\implies\tan^2\theta-\sin^2\theta$

$\implies\left(\dfrac{\sin\theta}{\cos\theta}\right)^2-\sin^2\theta$

$\implies\dfrac{\sin^2\theta}{\cos^2\theta}-\sin^2\theta$

Taking sin²θ common,

$\implies\sin^2\theta\left(\dfrac{1}{\cos^2\theta}-1\right)$

Taking LCM,

$\implies\sin^2\theta\left(\dfrac{1-\cos^2\theta}{\cos^2\theta}\right)$

We know that,

$\hookrightarrow 1-\cos^2 A=\sin^2 A$

Hence,

$\implies\sin^2\theta\left(\dfrac{\sin^2\theta}{\cos^2\theta}\right)$

We also know that,

$\hookrightarrow \dfrac{1}{\cos A}=\sec A$

So,

$\implies\sin^2\theta\left(\sin^2\theta\sec^2\theta\right)$

Hence,

$\implies\bf sin^4\theta sec^2\theta = RHS$

As, LHS = RHS

HENCE PROVED!!!