Question

prove that root of secant squared theta + cosec and square theta is equal to tan theta + cot theta ​

Answer 1

To Prove

$\sf\sqrt{sec^2\theta + cosec^2\theta} = tan\theta + cot\theta$

Solution

Taking LHS,

We know that

sec²∅ = tan²∅ + 1

and

cosec²∅ = cot²∅ + 1

So, replacing sec²∅ and cosec²∅, we get

$\sf\sqrt{tan^2\theta + 1 + cot^2\theta + 1}$

$\sf\sqrt{tan^2\theta + cot^2\theta + 2}$

Now,

2 can be written as 2 × 1

And, we know that 1 = tan∅cot∅

→ 2 = 2 tan∅cot∅

So we get

$\sf\sqrt{tan^2\theta + cot^2\theta + 2tan\theta cot\theta}$

Using a² + b² + 2ab = (a + b)²

$\sf\sqrt{(tan\theta + cot\theta)^2}$

= tan∅ + cot∅

Hence Proved