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Question: The value of \({{\tan }^{2}}\theta -{{\sec }^{2}}\theta \) is equal to...

The value of tan⁡2θ−sec⁡2θ{{\tan }^{2}}\theta -{{\sec }^{2}}\theta is equal to

Explanation

Solution

From the question given we have been asked to find the value of tan⁡2θ−sec⁡2θ{{\tan }^{2}}\theta -{{\sec }^{2}}\theta . To solve this question, we have to write the tan and sec in terms of sin and cos. As we know that tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta } and sec⁡θ=1cos⁡θ\sec \theta =\dfrac{1}{\cos \theta }. And also, we know the identity sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1. By using all these formulas, we will get the required answer.

Complete step by step solution:
From the question given we have to find the value of
⇒tan⁡2θ−sec⁡2θ\Rightarrow {{\tan }^{2}}\theta -{{\sec }^{2}}\theta
To solve this question, we have to write the tan and sec in terms of sin and cos.
As we know that the tan in terms of sin and cos is,
⇒tan⁡θ=sin⁡θcos⁡θ\Rightarrow \tan \theta =\dfrac{\sin \theta }{\cos \theta }
As we know that the sec in terms of cos is,
⇒sec⁡θ=1cos⁡θ\Rightarrow \sec \theta =\dfrac{1}{\cos \theta }
Now we have to substitute this in the given equation,
Now by substituting these in the given equation we will get,
⇒tan⁡2θ−sec⁡2θ=(sin⁡θcos⁡θ)2−(1cos⁡θ)2\Rightarrow {{\tan }^{2}}\theta -{{\sec }^{2}}\theta ={{\left( \dfrac{\sin \theta }{\cos \theta } \right)}^{2}}-{{\left( \dfrac{1}{\cos \theta } \right)}^{2}}
Now by further simplification we will get,
⇒tan⁡2θ−sec⁡2θ=sin⁡2θ−1cos⁡2θ\Rightarrow {{\tan }^{2}}\theta -{{\sec }^{2}}\theta =\dfrac{{{\sin }^{2}}\theta -1}{{{\cos }^{2}}\theta }
As we know the trigonometric identity that is,
⇒sin⁡2θ+cos⁡2θ=1\Rightarrow {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1
Rearranging this equation, we will get,
⇒sin⁡2θ−1=−cos⁡2θ\Rightarrow {{\sin }^{2}}\theta -1=-{{\cos }^{2}}\theta
Now substitute this in the above equation we will get,
⇒tan⁡2θ−sec⁡2θ=sin⁡2θ−1cos⁡2θ\Rightarrow {{\tan }^{2}}\theta -{{\sec }^{2}}\theta =\dfrac{{{\sin }^{2}}\theta -1}{{{\cos }^{2}}\theta }
⇒tan⁡2θ−sec⁡2θ=−cos⁡2θcos⁡2θ\Rightarrow {{\tan }^{2}}\theta -{{\sec }^{2}}\theta =\dfrac{-{{\cos }^{2}}\theta }{{{\cos }^{2}}\theta }
Now by further simplification we will get,
⇒tan⁡2θ−sec⁡2θ=−1\Rightarrow {{\tan }^{2}}\theta -{{\sec }^{2}}\theta =-1
Therefore, the value of tan⁡2θ−sec⁡2θ{{\tan }^{2}}\theta -{{\sec }^{2}}\theta is equal to −1-1.

Note: Students should recall all the formulas of trigonometry before doing this problem, student should know some trigonometric identities like sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1, sec⁡2θ−tan⁡2θ=1{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1, cos⁡ec2θ−cot⁡2θ=1\cos e{{c}^{2}}\theta -{{\cot }^{2}}\theta =1.