Question

if sec theta+tan theta =p ,prove that cos square theta
= (p square- 1) / (p square + 1)

Answer 1

$\huge{\underline{\underline{\sf{Answer}}}}$

Given that sec∅ + tan∅ = p.........(1)

To prove:

$\sf{sin \theta \: = \frac{p {}^{2} - 1 }{p {}^{2} + 1} }$

Here,

$\sf{ \frac{1}{p} = \frac{1}{sec \theta \: - \: tan \theta} } \\ \\ \implies \: \sf{ \frac{1}{p} = \frac{sec \theta \: - tan \theta}{sec {}^{2} \theta \: - \: tan {}^{2} \theta } } \\ \\ \sf{since \: \: sec {}^{2}x - tan {}^{2}x = 1 } \\ \\ \implies \: \boxed{\sf{ \frac{1}{p} = sec \theta \: - tan \theta}}.........(2)$

•Subtracting equations (1) and (2),

$\sf{p - \frac{1}{p} = (sec \theta + tan \theta) - (sec \theta \: - tan \theta) } \\ \\ \rightarrow \: \boxed{ \sf{ \frac{p {}^{2} - 1}{p} = 2tan \theta}}........(3)$

•Adding equations (1) and (2),we get:

$\sf{p + \frac{1}{p} = (sec \theta + \cancel{tan \theta}) + (sec \theta \: - \cancel{tan \theta}) } \\ \\ \rightarrow \: \boxed{ \sf{ \frac{p {}^{2} + 1}{p} = 2sec \theta}}...........(4)$

•Dividing equations (3) and (4),we get:

$\implies \: \sf{ \frac{p {}^{2} - 1 }{p {}^{2} + 1 } = \frac{2tan \theta}{2sec \theta} } \\ \\ \sf{sin \theta = \frac{tan \theta}{sec \theta}} \\ \\ \ \implies \: \underline{\underline{\boxed{\sf{ \frac{p {}^{2} - 1 }{p {}^{2} + 1 } = sin \theta}}}}$

Hence,proved

Answer 2

Correct Question :

If secØ + tanØ =p, prove that sinØ = (p² - 1)/(p² + 1)

Solution :

$\implies\:sec\theta\:+\:tan\theta\:=\:p$ ____ (eq 1)

We know that..

→ sec²Ø - tan²Ø = 1

So,

$\implies\:(sec\theta\:-\:tan\theta)(sec\theta\:+\:tan\theta)\:=\:1$

But secØ + tanØ = p

$\implies\:p\:(sec\theta\:-\:tan\theta)\:=\:1$

$\implies\:sec\theta\:-\:tan\theta\:=\:\dfrac{1}{p}$

$\implies\:sec\theta\:-\:tan\theta\:=\:\dfrac{1}{p}$ ____ (eq 2)

Now, add (eq 1) and (eq 2)

$\Rightarrow\:sec\theta\:+\:tan\theta\:+\:sec\theta\:-\:tan\theta\:=\:p\:+\:\dfrac{1}{p}$

$\Rightarrow\:2sec\theta\:=\:p\:+\:\dfrac{1}{p}$

$\Rightarrow\:\:2tan\theta\:=\: \dfrac{ {p}^{2} \: + \: 1 }{p}$ ____ (eq 3)

Now, subtract (eq 1) and (eq 2)

$\Rightarrow\:sec\theta\:+\:tan\theta\:-\:sec\theta\:+\:tan\theta\:=\:p\:-\:\dfrac{1}{p}$

$\Rightarrow\:2tan\theta\:=\:p\:-\:\dfrac{1}{p}$

$\Rightarrow\:2tan\theta\:=\: \dfrac{ {p}^{2} \: - \: 1 }{p}$ _____ (eq 4)

Divide (eq 4) by (eq 3)

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

Now,

tanØ = sinØ/cosØ and secØ = 1/coØ

$\Rightarrow \: \dfrac{ \frac{sin \theta}{cos \theta} }{ \frac{1}{cos \theta} } \: = \: \dfrac{ \frac{ {p}^{2} \: - \: 1}{p} }{ \frac{ {p}^{2} \: + \: 1}{p} }$

$\Rightarrow\:sin\theta\:=\:\dfrac{ {p}^{2} \: - \: 1}{ {p}^{2} \: + \: 1 }$

Hence, proved.