Question

if sec theat + tan theat =p prove that sin theat =(p²-1)/(p²+1)​

Answer 1

Solution:

We have ,

sec\theta + tan\theta = p ----(1)

Now,

sec^{2}\theta-tan^{2}\theta = 1

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

\implies p\times (sec\theta - tan\theta)=1

\implies sec\theta + tan\theta = \frac{1}{p}--(2)

Adding and subtracting (1) and (2) , we get

2sec\theta = \frac{(p^{2}+1)}{p} ---(3)

2tan\theta = \frac{(p^{2}-1)}{p} ---(4)

on dividing equation (4) by (3), we get

\frac{2tan\theta}{2sec\theta}=\frac{\frac{(p^{2}-1)}{p}}{\frac{(p^{2}+1)}{p}}

\implies \frac{\frac{sin\theta}{cos\theta}}{\frac{1}{cos\theta}}= \frac{\frac{(p^{2}-1)}{p}}{\frac{(p^{2}+1)}{p}}

After cancellation, we get

sin\theta = \frac{\frac{(p^{2}-1)}{p}}{\frac{(p^{2}+1)}{p}}

Hence, proved .

••••....

hope it will help you.....

Answer 2

Answer:

$\sin \alpha = \frac{ {p}^{2} - 1 }{ {p}^{2} + 1 }$

$= \frac{( \sec \alpha + tan \alpha ) ^{2} - 1 }{ {(sec } \alpha +tan \alpha ) ^{2} + 1 } \\ \frac{ {sec}^{2} \alpha + {tan}^{2} \alpha + 2tan \alpha sec \alpha - 1}{ {sec}^{2} \alpha + {tan}^{2} \alpha + 2tan \alpha sec \alpha + 1 } \\ \frac{tan ^{2} \alpha + {tan}^{2} \alpha + 2tan \alpha sec \alpha }{ {sec \alpha }^{2} + {sec}^{2} \alpha + 2tan \alpha sec \alpha } \\ \frac{2 {tan}^{2} \alpha + 2tan \alpha sec \alpha }{2 {sec}^{2} \alpha - + 2tan \alpha sec \alpha } \\ \frac{tan \alpha (2tan \alpha + 2sec \alpha) }{sec \alpha (2tan \alpha + 2sec \alpha )} \\ \frac{tan \alpha }{ \sec \alpha } \\ \frac{ \frac{sin \alpha }{ \cos \alpha } }{ \frac{1}{cos \alpha } } = sin \alpha$