Question

Sin 90 minus theta into cos theta + cos 90 minus theta into sin theta upon tan theta into cot theta

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

$\frac{\sin(90-\theta)\times \cos \theta+\cos(90-\theta)\times \sin \theta}{\tan\theta\times \cot\theta}=1$

Step-by-step explanation:

Given : Expression 'Sin 90 minus theta into cos theta + cos 90 minus theta into sin theta upon tan theta into cot theta'.

To find : The solution of the given expression?

Solution :

We can re-write the given expression as

$\frac{\sin(90-\theta)\times \cos \theta+\cos(90-\theta)\times \sin \theta}{\tan\theta\times \cot\theta}$

Now, To solve the expression apply trigonometric identities

$\sin(90-\theta)=\cos\theta$ and $\cos(90-\theta)=\sin\theta$

Substitute in the expression,

$=\frac{\cos\theta\times \cos \theta+\sin\theta\times \sin \theta}{\tan\theta\times \cot\theta}$

$=\frac{\cos^2\theta+\sin^2\theta}{\tan\theta\times \cot\theta}$

We know, $\cos^2\theta+\sin^2\theta=1$ and $\cot\theta=\frac{1}{\tan\theta}$

$=\frac{1}{\tan\theta\times\frac{1}{\tan\theta}}$

$=\frac{1}{1}$

$=1$

So, $\frac{\sin(90-\theta)\times \cos \theta+\cos(90-\theta)\times \sin \theta}{\tan\theta\times \cot\theta}=1$