Question

Prove that : \[{(\sin \theta + cosec\theta )^2} + {(\cos \theta + \sec \theta )^2} = 7 + {\tan ^2}\theta + {\cot ^2}\theta \] .

Answer 1

Hint: This is a trigonometry based question. You should have a good grasp on trigonometric ratios to solve these types of questions. It may seem to be scary when you look at the question first, but you can solve it very smoothly, once you understand trigonometric ratios completely. Basically, there are six trigonometric ratios, as you must already know, sin, cos, tan, cosec, cot and sec. Let us solve the problem.

Step wise solution:
Given data:
To prove \[{(\sin \theta + cosec\theta )^2} + {(\cos \theta + \sec \theta )^2} = 7 + {\tan ^2}\theta + {\cot ^2}\theta \]
We will take the LHS and prove that it is equal to the RHS.
\[LHS = {(\sin \theta + cosec\theta )^2} + {(\cos \theta + \sec \theta )^2}\]
After expanding the square, we get

\[LHS = {\sin ^2}\theta + {cosec^2}\theta + 2\sin \theta \cdot cosec\theta + {\cos ^2}\theta + {\sec ^2}\theta + 2\cos \theta \cdot \sec \theta \]
 \[ = ({\sin ^2}\theta + {\cos ^2}\theta ) + ({cosec^2}\theta + {\sec ^2}\theta ) + 2\sin \theta \cdot cosec\theta + 2\cos \theta \cdot \sec \theta \]


As we know that:

\[ {\sin ^2}\theta + {\cos ^2}\theta = 1 \\ \] and
\[\cos \theta = \dfrac{1}{{\sec \theta }}\\ \]
i.e.:\[ \,\cos \theta \cdot \sec \theta = 1\\ \]
Also\[ \,cosec\theta = \dfrac{1}{{\sin \theta }}\\ \]
 \[ \Rightarrow \sin \theta \cdot cosec\theta = 1 \]

Then, LHS can be written as,

LHS $ = 1 + \cos e{c^2}\theta + {\sec ^2}\theta + 2 \times 1 + 2 \times 1\\
   = (1 + 2 + 2) + \cos e{c^2}\theta + {\sec ^2}\theta $

Again, we know that ,

\[ \cos e{c^2}\theta - {\cot ^2}\theta = 1\\ \]
i.e.:\[ \cos e{c^2}\theta = 1 + {\cot ^2}\theta \\ \]
Also,\[ {\sec ^2}\theta = 1 + {\tan ^2}\theta \\ \]

Then, we can put the values of the \[\cos e{c^2}\theta \,\,and\,{\sec ^2}\theta \] in LHS, and we obtain,

\[ LHS = (1 + 2 + 2) + (1 + {\cot ^2}\theta ) + (1 + {\tan ^2}\theta )\\ \]
   \[ = (1 + 2 + 2 + 1 + 1) + {\cot ^2}\theta + {\tan ^2}\theta \\ \]

Add the values inside the bracket

$ LHS = 7 + {\cot ^2}\theta + {\tan ^2}\theta \\
   = 7 + {\tan ^2}\theta + {\cot ^2}\theta \\
   = RHS $

i.e.: LHS = RHS

Note: You should be very careful with the relations of trigonometric ratios with each other. Students often make mistakes in putting the relations in sum and computing.