Question

Sec thetha + tan thetha = p prove that p2-1/p2+1= sin thetha

Answer 1

Answer:

Answer

p2+1p2−1

=(secθ+tanθ)2+1(secθ+tanθ)2−1

sec2θ+tan2θ+2secθtanθ+1sec2θ+tan2θ+2secθtanθ−1

=sec2θ+(tan2θ+1)+2secθtanθ(sec2θ−1)+tan2θ+2secθtanθ

=sec2θ+sec2θ+2secθtanθtan2θ+tan2θ+2secθtanθ

=2sec2θ+2secθtanθ2tan2θ+2secθtanθ

=

Answer 2

$\huge{\fbox{\fcolorbox{cyan}{pink}{✿AnsWer}}}</p><p></p><h3>● GIVEN:-</h3><p></p><p>[tex]{Sec\theta + tan\theta =p}$

●TO PROVE:-

\bold{ \frac{ {p}^{2} - 1 }{ {p}^{2} + 1 } = sin \theta }[/tex]

● LET'S DO IT:-

----LHS

${\boxed{ \boxed {\boxed{ \bold{ \purple{ : \longrightarrow \: \frac{ {p}^{2} - 1}{ {p}^{2} + 1 } }}}}}}$

$= \frac{ {sec}^{2} \theta \: + {tan}^{2} \theta \: + 2sec \theta \: tan \: \theta - 1 }{ {sec}^{2}theta \: + {tan}^{2} theta \: + 2sec \: \theta \: tan \theta}$

$= \frac{ {tan}^{2} \theta \: + {tan}^{2} \theta \: + 2sec \theta \: tan \theta }{ {sec}^{2} \theta \: + {sec}^{2} \theta + 2sec \theta \: tan \theta}$

$= \frac{2 {tan}^{2} \theta \: + 2sec \theta \: tan \theta }{2 {sec}^{2} \theta \: + 2sec \: \theta \: tan \: \theta}$

$= \frac{2( \tan \theta)( \cancel{tan \theta \: + sec \theta})}{2sec \theta \: ( \cancel{sec \: \theta \: + tan \theta})}$

$= \blue{ sin \theta}$

----RHS

HOPE IT HELPS

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