Question

If sec theta + tan theta = p, prove that sin theta = p 2 -1 / p 2 + 1

Answer 1

p = tanФ + secФ
p - secФ = tanФ
p² + sec²Ф  - 2 p secФ = tan²Ф = sec²Ф - 1
So secФ = (p² + 1) / 2p

so cosФ = 2p/(1+p²)
sinФ = √[1 - cos²Ф ] = (p²-1)/(1+p²)

Answer 2

To prove:

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

Given:

$\sec \theta+\tan \theta=p$

Solution:

$p=\sec \theta+\tan \theta$

Hence, we can say that  

$p-\sec \theta=\tan \theta$

Squaring both sides, we get

$p^{2}+\sec ^{2} \theta-2 p \sec \theta=\tan ^{2} \theta$

$p^{2}+\sec ^{2} \theta-2 p \sec \theta=\sec ^{2} \theta-1$

$p^{2}+1-2 p \sec \theta=0$

$p^{2}+1=2 p \sec \theta$

So, $\sec \theta=\frac{p^{2}+1}{2 p}$

$\cos \theta=\frac{2 p}{1+p^{2}}$

$\sin \theta=\sqrt{1-\cos ^{2} \theta}=\frac{p^{2}-1}{p^{2}+1}$

Hence, proved that $\bold{\sin \theta=\frac{p^{2}-1}{p^{2}+1}}$ given that $\bold{\sec \theta+\tan \theta=p}$.

The ‘trigonometric functions’ also called ‘circular functions’, ‘angle functions’ or ‘goniometricfunctions’ are “real functions” which relate an ‘angle’ of a ‘right-angled triangle’ to ratios of two side lengths.