Question

GIVEN,
cosec thetha + cot thetha = p

prove, p^2 -1
---------- = cos thetha
p^2 +1 ​

Answer 1

Given

cosec∅ + cot∅ = p ----> (1)

We know that

cosec²∅ = 1 + cot²∅

=> cosec²∅ - cot²∅ = 1

Using Identity a² - b² = (a+b) (a-b)

=> (cosec∅ + cot∅)(cosec∅ - cot∅) = 1

=> p (cosec∅ - cot∅) = 1

=> cosec∅ - cot∅ = 1/p ----> (2)

Adding (1) and (2) we get:

2cosec∅ = $\frac{p^2 + 1}{p}$

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

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

We know that, cos∅ = $\sqrt{1 - sin^2\theta}$

=>cos∅ $= \sqrt{1 - (\frac{2p}{p^2+1})^2}$

= $\sqrt{\frac{p^4 + 1 + 2p^2 - 4p^2}{(p^2 + 1)^2}}$

= $\sqrt{\frac{p^4 + 1 - 2p^2}{(p^2 + 1)^2}}$

= $\sqrt{\frac{(p^2 - 1)^2}{(p^2 + 1)^2}}$

= > cos∅ = $\frac{p^2 - 1}{p^2 + 1}$

Hence proved

Answer 2

Answer:

The following pictures have the solution.