Question

twn square theta upon sec theta +1 = sec theta - 1 prove this​

Answer 1

The basic way to prove this kind of questions is to match the LHS and RHS of the given equation :

LHS:

$\therefore \dfrac{ { \tan}^{2}( \theta )}{ \sec( \theta) + 1 }$

$= \dfrac{ { \sec}^{2}( \theta) - 1 }{ \sec( \theta) + 1}$

$= \dfrac{ { \sec}^{2}( \theta) - {(1)}^{2} }{ \sec( \theta) + 1}$

$= \dfrac{ \{\sec( \theta) + 1 \} \{ \sec( \theta) - 1 \} }{ \{ \sec( \theta) + 1 \}}$

Cancelling similar terms , we get :

$= \dfrac{ \cancel{\{\sec( \theta) + 1 \}} \{ \sec( \theta) - 1 \} }{ \cancel{\{ \sec( \theta) + 1 \}}}$

$= \sec( \theta) - 1$

RHS :

$= \sec( \theta) - 1$

Hence , LHS = RHS

Basic identities used :

$1) \: {a}^{2} - {b}^{2} = (a + b)(a - b)$

$2) \: { \tan}^{2} ( \theta) + 1 = { \sec}^{2} ( \theta)$

Answer 2

$\sf{\huge{\mathbb{\pink{\bigstar{\boxed{\boxed{Question}}}}}}}$

prove :-

$\sf\dfrac{tan^2\theta}{sec\theta + 1 } = sec\theta - 1$

$\sf{\huge{\mathbb{\pink{\bigstar{\boxed{\boxed{Solution}}}}}}}$

$\implies\sf\dfrac{tan^2\theta}{sec\theta + 1 } = sec\theta - 1$

$\sf{\bold{\red{\boxed{sec^2\theta - tan^2\theta = 1 }}}}$

$\sf{\bold{\red{\boxed{tan^2 \theta = 1 - sec^2\theta}}}}$

$\implies\sf\dfrac{sec^2\theta - 1^2 }{sec\theta + 1 } = sec\theta - 1$

$\sf{\bold{\red{\boxed{a^2 - b^2 = (a+ b)(a - b)}}}}$

$\implies\sf\dfrac{(sec\theta - 1)\cancel{(sec\theta + 1)}}{\cancel{sec\theta + 1 }} = sec\theta - 1$

$\implies\sf sec\theta - 1 = sec\theta - 1$

$\sf{\bold{\orange{\boxed{\dag LHS = RHS}}}}$

$\sf{\bold{\orange{\boxed{\dag Hence\:Proved}}}}$

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