Question

If $cosec\theta +\cot \theta =p$, then prove that $\cos \theta =\frac{p^2-1}{p^2+1}$.

Answer 1

given $co\sec \theta +\cot \theta =P...\left(1\right)$

WE know that $1+{\cot }^2\theta =co{\sec }^2\theta \Rightarrow co{\sec }^2\theta -{\cot }^2\theta =1$

$\Rightarrow (co\sec \theta -\cot \theta )(co\sec \theta +\cot \theta )=1\Rightarrow (co\sec \theta -\cot \theta )P=1$

$\Rightarrow co\sec \theta -\cot \theta =\frac{1}{P}...\left(2\right)$

Adding (1) and (2)

$o\sec \theta =\frac{P+\frac{1}{P}}{2};\cot \theta =\frac{P-\frac{1}{P}}{2}$

$\cos \theta =\frac{\cot \theta }{co\sec \theta }=\frac{\frac{P-\frac{1}{P}}{2}}{\frac{P+\frac{1}{P}}{2}}=\frac{P^2-1}{P^2+1}$

Hencd proved