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Question: Prove the given trigonometric expression. The trigonometric expression is \[se{{c}^{2}}\theta +co{{t...

Prove the given trigonometric expression. The trigonometric expression is sec2θ+cot2(90−θ)=2  cosec2(90−θ)−1se{{c}^{2}}\theta +co{{t}^{2}}\left( 90-\theta \right)=2\;cose{{c}^{2}}\left( 90-\theta \right)-1

Explanation

Solution

Hint: First, simplify the expression in the LHS sec2θ+cot2(90−θ)se{{c}^{2}}\theta +co{{t}^{2}}\left( 90-\theta \right)using the identity cot(90−θ)=tan⁡θcot\left( 90-\theta \right)=\tan \theta and sec⁡2(θ)−tan⁡2(θ)=1{{\sec }^{2}}\left( \theta \right)-{{\tan }^{2}}\left( \theta \right)=1. Next simplify the expression is the RHS 2  cosec2(90−θ)−12\;cose{{c}^{2}}\left( 90-\theta \right)-1using the identity cosec(90−θ)=sec⁡θcosec\left( 90-\theta \right)=\sec \theta . We should get LHS =RHS in order to prove the above result.

Complete step-by-step answer:
In the question, we have to prove that sec2θ+cot2(90−θ)=2  cosec2(90−θ)−1se{{c}^{2}}\theta +co{{t}^{2}}\left( 90-\theta \right)=2\;cose{{c}^{2}}\left( 90-\theta \right)-1.
So in order to prove that we have just show that trigonometric expression in the Left hand side (LHS) is equal to the trigonometric expression in the right hand side (RHS).
So here we will start with the LHS and then simplify it, as follows:

& \Rightarrow LHS=se{{c}^{2}}\theta +co{{t}^{2}}\left( 90-\theta \right) \\\ & \Rightarrow LHS=se{{c}^{2}}\theta +{{\tan }^{2}}\left( \theta \right)\,\,\,\,\,\,\,\,\because cot\left( 90-\theta \right)=\tan \left( \theta \right)\, \\\ \end{aligned}$$ So here we have used the identity $$cot\left( 90-\theta \right)=\tan \left( \theta \right)\,$$. Next, we will further write this LHS as: $$\begin{aligned} & \Rightarrow LHS=se{{c}^{2}}\theta +{{\tan }^{2}}\left( \theta \right)\, \\\ & \Rightarrow LHS=2se{{c}^{2}}\theta -1\,\,\,\,\,\,\,\,\,\,\because {{\tan }^{2}}\left( \theta \right)={{\sec }^{2}}\left( \theta \right)-1 \\\ \end{aligned}$$ Next, we will take the RHS, and simplify it as follows: $$\begin{aligned} & \Rightarrow RHS=2\;cose{{c}^{2}}\left( 90-\theta \right)-1 \\\ & \Rightarrow RHS=2\;{{\sec }^{2}}\left( \theta \right)-1\,\,\,\,\,\,\,\,\,\,\because cosec\left( 90-\theta \right)=\;\sec \left( \theta \right) \\\ \end{aligned}$$ So here we see that $$LHS=2se{{c}^{2}}\theta -1$$ and $$RHS=2\;{{\sec }^{2}}\left( \theta \right)-1\,\,$$, which shows that LHS=RHS and this is what we have to prove. Hence, we have proven that $$se{{c}^{2}}\theta +co{{t}^{2}}\left( 90-\theta \right)=2\;cose{{c}^{2}}\left( 90-\theta \right)-1$$ Note: It is important that we apply the correct identity. For example, $${{\sec }^{2}}\left( \theta \right)-{{\tan }^{2}}\left( \theta \right)=1$$ and not $${{\sec }^{2}}\left( \theta \right)+{{\tan }^{2}}\left( \theta \right)\ne 1$$, also we have $$cosec\left( 90-\theta \right)=\sec \theta $$ and not $$cosec\left( 90-\theta \right)\ne -\sec \theta $$. The other method that can be used here is to bring all the trigonometric expressions in LHS and make RHS=0. Then simplify the LHS to get zero in order to prove that the Left hand side (LHS) is equal to the right hand side (RHS).