Question

Show that \[tan^{2}\theta-\dfrac{1}{\cos^{2}\theta} = - 1\]

Answer 1

Hint: In this question, we need to prove that \[tan^{2}\theta-\dfrac{1}{\cos^{2}\theta}\] is equal to \[- 1\] . In order to prove this equation, we need to know the trigonometric identities. Then we can expand the right side of the expression which was given to prove to get the left side of the expression. By using trigonometric identities and functions, we can prove this.

Complete step by step answer:
First we can consider the left side of the given equation,
⇒ \[tan^{2}\theta-\dfrac{1}{\cos^{2}\theta}\]
We know that \[tan\theta = \dfrac{{sin\theta}}{{cos\theta}}\]
By squaring both sides,
We get,
\[\tan^{2}\theta = \dfrac{\operatorname{sin^{2}}\theta}{\cos^{2}\theta}\]
By substituting in the given equation,
We get,
⇒ \[\dfrac{\operatorname{sin^{2}}\theta}{\cos^{2}\theta} - \dfrac{1}{\cos^{2}\theta}\]
⇒ \[\dfrac{sin^{2}\theta – 1}{{co}s^{2}\theta}\]
We also know that \[sin^{2}\theta + {co}s^{2}\theta = 1\]
From this we get,
\[sin^{2}\theta – 1 = - {co}s^{2}\theta\]
By substituting this,
We get,
⇒ \[\dfrac{-cos^{2}\theta}{{co}s^{2}\theta}\]
By dividing,
We get,
⇒ \[- 1\]
Thus we get the right side of the equation.
Therefore we have proved \[tan^{2}\theta-\dfrac{1}{\cos^{2}\theta} = - 1\]

Note:
Alternative solution :
Consider the left side of the equation,
\[tan^{2}\ \theta-\dfrac{1}{cos^{2}\theta}\]
We know that \[\dfrac{1}{cos^{2}\theta} = sec^{2}\theta\]
By substituting this we get,
⇒ \[tan^{2}\ \theta-sec^{2}\theta\]
By taking the negative sign outside,
⇒ \[- \left( \left( - tan^{2}\theta\ \right) + sec^{2}\theta \right)\]
By rearranging the terms,
We get,
⇒ \[- (sec^{2}\theta – tan^{2}\theta)\]
We know that \[sec^{2}\theta – tan^{2}\theta = 1\]
Thus we get \[tan^{2}\theta-\dfrac{1}{\cos^{2}\theta} = - 1\]