Question

Find the value of the trigonometric expression $(1 + {\tan ^2}\theta ) \cdot {\sin ^2}\theta $
a. ${\sin ^2}\theta $
b. ${\cos ^2}\theta $
c. ${\tan ^2}\theta $
d. ${\cot ^2}\theta $

Answer 1

Hint: Trigonometric identities like sin²θ+cos²θ=1 can be used to rewrite expressions in a different, more convenient way. For example, $(1 - {\cos ^2}\theta )({\sin ^2}\theta )$ can be rewritten as $({\sin ^2}\theta )({\sin ^2}\theta )$ and then as $({\sin ^4}\theta )$. In this question, we can use trigonometric identity ${\cos ^2}\theta + {\sin ^2}\theta = 1$ and for alternative method $1 + {\tan ^2}\theta = {\sec ^2}\theta $.

Complete step-by-step answer:

Let us consider the trigonometric expression
$(1 + {\tan ^2}\theta ) \cdot {\sin ^2}\theta = \left[ {1 + {{(\tan \theta )}^2}} \right] \cdot {\sin ^2}\theta $

We know that the trigonometric function $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$, we get
\[(1 + {\tan ^2}\theta ) \cdot {\sin ^2}\theta = \left[ {1 + {{\left( {\dfrac{{\sin \theta }}{{\cos \theta }}} \right)}^2}} \right] \cdot {\sin ^2}\theta \]

By using the formula ${\left( {\dfrac{a}{b}} \right)^2} = \dfrac{{{a^2}}}{{{b^2}}}$ , then the left hand side is as follow
\[(1 + {\tan ^2}\theta ) \cdot {\sin ^2}\theta = \left[ {1 + \dfrac{{{{(\sin \theta )}^2}}}{{{{(\cos \theta )}^2}}}} \right] \cdot {\sin ^2}\theta \]

Now, the trigonometric functions are ${(\sin \theta )^2} = {\sin ^2}\theta $ and ${(\cos \theta )^2} = {\cos ^2}\theta $, then the trigonometric expression is given by
\[(1 + {\tan ^2}\theta ) \cdot {\sin ^2}\theta = \left[ {1 + \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right] \cdot {\sin ^2}\theta \]
\[(1 + {\tan ^2}\theta ) \cdot {\sin ^2}\theta = \left[ {\dfrac{{{{\cos }^2}\theta + {{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right] \cdot {\sin ^2}\theta \]

Applying the trigonometric identity ${\cos ^2}\theta + {\sin ^2}\theta = 1$, we get
\[(1 + {\tan ^2}\theta ) \cdot {\sin ^2}\theta = \left[ {\dfrac{1}{{{{\cos }^2}\theta }}} \right] \cdot {\sin ^2}\theta \]

The rearranging the terms, we get
\[(1 + {\tan ^2}\theta ) \cdot {\sin ^2}\theta = \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}\]

By using the formula $\dfrac{{{a^2}}}{{{b^2}}} = {\left( {\dfrac{a}{b}} \right)^2}$, then left hand side term of the trigonometric expression is as follow
\[(1 + {\tan ^2}\theta ) \cdot {\sin ^2}\theta = {\left( {\dfrac{{\sin \theta }}{{\cos \theta }}} \right)^2}\]

Again, we use the trigonometric function $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$.
\[(1 + {\tan ^2}\theta ) \cdot {\sin ^2}\theta = {\left( {\tan \theta } \right)^2}\]
The trigonometric functions ${(\tan \theta )^2}$ and ${\tan ^2}\theta $ both are the same.

Thus, the trigonometric expression is \[(1 + {\tan ^2}\theta ) \cdot {\sin ^2}\theta = {\tan ^2}\theta \].

Hence the correct option of the given question is option (c).

Note: Alternatively this question is solved as follows-
We know that the trigonometric second identity ${\sec ^2}\theta = 1 + {\tan ^2}\theta $, then the given trigonometric expression is $({\sec ^2}\theta ) \cdot {\sin ^2}\theta $.

Again, we use the trigonometric function ${\sec ^2}\theta = \dfrac{1}{{{{\cos }^2}\theta }}$ and then the given trigonometric expression is \[\left( {\dfrac{1}{{{{\cos }^2}\theta }}} \right) \cdot {\sin ^2}\theta \].

Finally, the trigonometric function \[\left( {\dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right)\] is ${\tan ^2}\theta $.

Hence the correct option of the given question is option (c).