Question

Sin²thetaCot²theta+Cos²thetaTan²theta=​

Answer 1

Answer:

1

Step-by-step explanation:

Given :

$\mathrm {sin^2\theta .cot^2\theta+ cos^2\theta .tan^2\theta}$

To find :

The value of the given expression

Solution :

We know that,

$\boxed{\mathrm {cotx = \dfrac{cosx}{sinx} \ and \ tanx = \dfrac{sinx}{cosx}}}$

Substituting in the expression

$\implies \mathrm{sin^2\theta \times \dfrac{cos^2\theta}{sin^2 \theta} + cos^2\theta \times \dfrac{sin^2\theta}{cos^2\theta}}$

Cancelling out the terms, we get,

$\implies \mathrm{\bcancel{sin^2\theta} \times \dfrac{cos^2\theta}{\bcancel{sin^2 \theta}} + \bcancel{cos^2\theta} \times \dfrac{sin^2\theta}{\bcancel{cos^2\theta}}}$

$\implies \mathrm {cos^2\theta + sin^2\theta}$

We know that,

$\boxed{\mathrm{cos^2x + sin^2x = 1}}$

Thus,

$\implies \huge{\texttt{\underline{\underline{1}}}}$

Hope it helps!!