Question

Prove with the help of trigonometric identities $2{\sin ^2}\theta + 4{\sec ^2}\theta + 5{\cot ^2}\theta + 2{\cos ^2}\theta - 4{\tan ^2}\theta - 5\cos e{c^2}\theta = 1$

Answer 1

Hint: Here in this question we will apply some trigonometric identities to solve and prove left hand side equal to right hand side.
${\sin ^2}\theta + {\cos ^2}\theta = 1$
${\sec ^2}\theta = {\tan ^2}\theta + 1$
$\cos e{c^2}\theta = 1 + {\cot ^2}\theta $

Complete step-by-step answer:
We will take the L.H.S part and will try to convert it into the R.H.S part.
$ \Rightarrow 2{\sin ^2}\theta + 4{\sec ^2}\theta + 5{\cot ^2}\theta + 2{\cos ^2}\theta - 4{\tan ^2}\theta - 5\cos e{c^2}\theta $
Rearranging the terms so that it becomes easy in applying suitable identities.
$ \Rightarrow 2{\sin ^2}\theta + 2{\cos ^2}\theta + 4{\sec ^2}\theta - 4{\tan ^2}\theta + 5{\cot ^2}\theta - 5\cos e{c^2}\theta $
Taking common alike terms
$ \Rightarrow 2({\sin ^2}\theta + {\cos ^2}\theta ) + 4({\sec ^2}\theta - {\tan ^2}\theta ) + 5({\cot ^2}\theta - \cos e{c^2}\theta )$
Applying the identities:-
${\sin ^2}\theta + {\cos ^2}\theta = 1$
\[{\sec ^2}\theta = {\tan ^2}\theta + 1\]
\[\therefore {\sec ^2}\theta - {\tan ^2}\theta = 1\]
$\cos e{c^2}\theta = 1 + {\cot ^2}\theta $
$\therefore {\cot ^2}\theta - \cos e{c^2}\theta = - 1$
$ \Rightarrow 2(1) + 4(1) + 5( - 1)$
$ \Rightarrow 2 + 4 - 5$
$ = 1$
$\therefore 2{\sin ^2}\theta + 4{\sec ^2}\theta + 5{\cot ^2}\theta + 2{\cos ^2}\theta - 4{\tan ^2}\theta - 5\cos e{c^2}\theta = 1$
Hence L.H.S=R.H.S proved.

Additional Information: Importance of trigonometric functions: - In geometry trigonometric functions are used to find unknown angles or sides of right angled triangles. The three common used trigonometric functions are $\sin \theta ,\cos \theta ,\tan \theta $.If we have good understanding of these three functions then other trigonometric functions such as $\cos ec\theta ,\sec \theta ,\cot \theta $ which are reciprocal of the above three functions can be easily understood.

Note: 1. This question can be done by various methods one method is by converting the whole equation into $\sin \theta $ and $\cos \theta $. But it will be very long and time consuming. The approach by which we have solved is way more convenient rather than converting the whole into $\sin \theta $ and $\cos \theta $ . So some identities need to be memorised so that it can be done easily and fast.
2. Presence of mind in these questions is very important because direct identity applying will not be asked so by rearrangement or by taking something common we have to make that identity so that we can use them. Also students should avoid unnecessary calculation mistakes so that it can be solved efficiently.