Question

How do you verify ${\tan ^2}(a) - {\sin ^2}(a) = {\tan ^2}(a){\sin ^2}(a)$ ?

Answer 1

Hint: Here, in this question, we are asked to verify the given trigonometric equation line. First, we will take any of the sides and substitute it with the help of properties or identity of the trigonometric functions and expand it as possible. We have to simplify it in such a way that it ends up resulting in the term on the other side.

Formula used: Trigonometric function identity:
${\sin ^2}x + {\cos ^2}x = 1$
\[{\tan ^2}(a) = \dfrac{{{{\sin }^2}(a)}}{{{{\cos }^2}(a)}}\]

Complete step-by-step solution:
The given trigonometric expression is ${\tan ^2}(a) - {\sin ^2}(a) = {\tan ^2}(a){\sin ^2}(a)$, we need to verify it.
First we will try to substitute the LHS terms with properties of identities of the trigonometric functions and expand it until we arrive at the RHS.
Taking the LHS, we have ${\tan ^2}(a) - {\sin ^2}(a)$,
We know that,\[{\tan ^2}(a) = \dfrac{{{{\sin }^2}(a)}}{{{{\cos }^2}(a)}}\], substituting it we get
$ \Rightarrow \dfrac{{{{\sin }^2}a}}{{{{\cos }^2}a}} - {\sin ^2}a$
Cross multiplying the denominator with the other term in order to make it as a one single term that is taking LCM. We get,
$ \Rightarrow \dfrac{{{{\sin }^2}a - {{\sin }^2}a{{\cos }^2}a}}{{{{\cos }^2}a}}$
Now, it is clear that, we have two \[{\sin ^2}(a)\] in the numerator and we can now take \[{\sin ^2}(a)\] as common out,
$ \Rightarrow \dfrac{{{{\sin }^2}a\left( {1 - {{\cos }^2}a} \right)}}{{{{\cos }^2}a}}$
We know that ${\sin ^2}x + {\cos ^2}x = 1$ and that implies that ${\sin ^2}x = 1 - {\cos ^2}x$ , so now we can replace it below,
\[ \Rightarrow \dfrac{{{{\sin }^2}a{{\sin }^2}a}}{{{{\cos }^2}a}}\]
Now that we have this, we can separate the term into two,
\[ \Rightarrow \dfrac{{{{\sin }^2}a}}{{{{\cos }^2}a}} \times {\sin ^2}a\]
We know that \[{\tan ^2}(a) = \dfrac{{{{\sin }^2}(a)}}{{{{\cos }^2}(a)}}\],
$ \Rightarrow {\tan ^2}a{\sin ^2}a$
= RHS
Hence the given expression is verified.

Note: In this question we have alternative method as follows:
We can verify the same using another method too.
$ \Rightarrow {\tan ^2}(a) - {\sin ^2}(a) = {\tan ^2}(a){\sin ^2}(a)$
Taking the right hand side, we have
$ \Rightarrow {\tan ^2}(a){\sin ^2}(a)$
We know that${\tan ^2}x = {\sec ^2}a - 1$, replacing the same we get
$ \Rightarrow ({\sec ^2}a - 1)({\sin ^2}a)$
Now we also know that, \[{\sec ^2}(a) = \dfrac{1}{{{{\cos }^2}(a)}}\] substituting it we get
$ \Rightarrow \left( {\dfrac{1}{{{{\cos }^2}a}} - 1} \right)\left( {{{\sin }^2}a} \right)$
Multiplying the \[{\sin ^2}(a)\] inside the brackets we get
$ \Rightarrow \left( {\dfrac{1}{{{{\cos }^2}a}}} \right)\left( {{{\sin }^2}a} \right) - 1({\sin ^2}a)$
$ \Rightarrow \left( {\dfrac{{{{\sin }^2}a}}{{{{\cos }^2}a}}} \right) - ({\sin ^2}a)$
We know that \[{\tan ^2}(a) = \dfrac{{{{\sin }^2}(a)}}{{{{\cos }^2}(a)}}\]
$ \Rightarrow {\tan ^2}a - {\sin ^2}a$
= LHS
Hence the given expression is verified.