Question

Prove the following trigonometric equation:
\[{{\tan }^{2}}a-{{\tan }^{2}}b=\dfrac{{{\sin }^{2}}a-{{\sin }^{2}}b}{{{\cos }^{2}}a.{{\cos }^{2}}b}\]

Answer 1

Hint: First of all, consider the LHS of the equation given in the question. Now use \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\] to write the expression for $\tan a$ and $\tan b$. Now, simplify the expression and use \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] to prove the desired result.

Complete step-by-step solution -
In this question, we have to prove that
\[{{\tan }^{2}}a-{{\tan }^{2}}b=\dfrac{{{\sin }^{2}}a-{{\sin }^{2}}b}{{{\cos }^{2}}a.{{\cos }^{2}}b}\]
Let us consider the LHS of the equation given in the question.
\[LHS={{\tan }^{2}}a-{{\tan }^{2}}b.....\left( i \right)\]
We know that \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\]. So, by taking \[\theta =a\], we get, \[\tan a=\dfrac{\sin a}{\cos a}\] and by taking \[\theta =b\], we get \[\tan b=\dfrac{\sin b}{\cos b}\]. So, by substituting the value of tan a and tan b in terms of sin a, cos a, sin b and cos b in equation (i), we get,
\[LHS={{\left( \dfrac{\sin a}{\cos a} \right)}^{2}}-{{\left( \dfrac{\sin b}{\cos b} \right)}^{2}}\]
We can also write the above equation as,
\[LHS=\dfrac{{{\sin }^{2}}a}{{{\cos }^{2}}a}-\dfrac{{{\sin }^{2}}b}{{{\cos }^{2}}b}\]
By taking \[{{\cos }^{2}}a{{\cos }^{2}}b\] as LCM and simplifying the above equation, we get,
\[LHS=\dfrac{{{\sin }^{2}}a{{\cos }^{2}}b-{{\sin }^{2}}b{{\cos }^{2}}a}{{{\cos }^{2}}a{{\cos }^{2}}b}\]
We know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\text{ or }{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \].
So, by taking \[\theta =a\], we get,
\[{{\cos }^{2}}a=1-{{\sin }^{2}}a\]
By taking \[\theta =b\], we get,
\[{{\cos }^{2}}b=1-{{\sin }^{2}}b\]
By using these in the above equation, we get,
\[LHS=\dfrac{{{\sin }^{2}}a\left( 1-{{\sin }^{2}}b \right)-{{\sin }^{2}}b\left( 1-{{\sin }^{2}}a \right)}{{{\cos }^{2}}a{{\cos }^{2}}b}\]
\[LHS=\dfrac{{{\sin }^{2}}a-{{\sin }^{2}}a{{\sin }^{2}}b-{{\sin }^{2}}b+{{\sin }^{2}}b{{\sin }^{2}}a}{{{\cos }^{2}}a.{{\cos }^{2}}b}\]
By canceling the like terms from the numerator of the above equation, we get,
\[LHS=\dfrac{{{\sin }^{2}}a-{{\sin }^{2}}b}{{{\cos }^{2}}a.{{\cos }^{2}}b}=RHS\]
So, we get, LHS = RHS
Hence, we have proved that
\[{{\tan }^{2}}a-{{\tan }^{2}}b=\dfrac{{{\sin }^{2}}a-{{\sin }^{2}}b}{{{\cos }^{2}}a.{{\cos }^{2}}b}\].

Note: In this question, we have changed cos a and cos b not sin a and sin b because, in the RHS of the equation, sin was present in the numerator. So, in these types of questions, students must follow the steps by keeping the other side of the equation in the mind because in trigonometry a single expression can be converted into multiple other expressions and not just one given on the other side of the equation by using different formulas. But we only have to follow those steps which result in LHS = RHS.