
Prove that ${\tan ^2}\theta {\cos ^2}\theta = 1 - {\cos ^2}\theta $
Answer
544.2k+ views
Hint:In order to prove the given equation, we need to use the trigonometric ratio $\tan \theta = \dfrac{{\sin \theta }} {{\cos \theta }}$. Then, we need to use the trigonometric identity ${\sin ^2}\theta = 1 - {\cos ^2}\theta $ to simplify the given expression. We can prove the given expression by showing that the left hand side is equal to the right hand side.
Complete step by step solution:
The given expression which is to be proved is ${\tan ^2}\theta {\cos ^2}\theta = 1 - {\cos ^2}\theta $ .
In the above equation, the left hand side is ${\tan ^2}\theta {\cos ^2}\theta $
and the right hand side is $1 - {\cos ^2}\theta $ .
To prove that the given expression is true, we need to show that the left hand side of the above equation is equal to the right hand side of the above equation that is, LHS = RHS.
It is known that trigonometric ratio of $\tan \theta $
is, $\tan \theta = \dfrac{{\sin \theta }}
{{\cos \theta }}$.
Substitute $\dfrac{{\sin \theta }}
{{\cos \theta }}$
for $\tan \theta $
in the left hand side of the given equation that is, ${\tan ^2}\theta {\cos ^2}\theta $ .
${\tan ^2}\theta {\cos ^2}\theta = \dfrac{{{{\sin }^2}\theta }}
{{{{\cos }^2}\theta }} \cdot {\cos ^2}\theta $
Cancel the numerator and denominator of the above expression that is, ${\cos ^2}\theta $
and ${\cos ^2}\theta $.
$\dfrac{{{{\sin }^2}\theta }}
{{{{\cos }^2}\theta }} \cdot {\cos ^2}\theta = {\sin ^2}\theta $
We know the trigonometric identity as, ${\sin ^2}\theta + {\cos ^2}\theta = 1$ .
We can rewrite the above identity in terms of ${\sin ^2}\theta $ as,
$
{\sin ^2}\theta + {\cos ^2}\theta = 1 \\
{\sin ^2}\theta = 1 - {\cos ^2}\theta \\
$
Substitute $1 - {\cos ^2}\theta $
for ${\sin ^2}\theta $
in the expression $\dfrac{{{{\sin }^2}\theta }}
{{{{\cos }^2}\theta }} \cdot {\cos ^2}\theta = {\sin ^2}\theta $.
$
\dfrac{{{{\sin }^2}\theta }}
{{{{\cos }^2}\theta }} \cdot {\cos ^2}\theta = {\sin ^2}\theta \\
= 1 - {\cos ^2}\theta \\
$
So we can write that ${\tan ^2}\theta {\cos ^2}\theta = 1 - {\cos ^2}\theta $ .
Thus, we get that the left hand side is equal to the right hand side that is, LHS = RHS.
Hence, the given statement is proved.
Note: To prove the given expression we must use the necessary trigonometric identities and ratios to simplify the expression. We can also prove the given expression by showing that the right hand side is equal to the left hand side that is, $1 - {\cos ^2}\theta = {\tan ^2}\theta {\cos ^2}\theta $
Complete step by step solution:
The given expression which is to be proved is ${\tan ^2}\theta {\cos ^2}\theta = 1 - {\cos ^2}\theta $ .
In the above equation, the left hand side is ${\tan ^2}\theta {\cos ^2}\theta $
and the right hand side is $1 - {\cos ^2}\theta $ .
To prove that the given expression is true, we need to show that the left hand side of the above equation is equal to the right hand side of the above equation that is, LHS = RHS.
It is known that trigonometric ratio of $\tan \theta $
is, $\tan \theta = \dfrac{{\sin \theta }}
{{\cos \theta }}$.
Substitute $\dfrac{{\sin \theta }}
{{\cos \theta }}$
for $\tan \theta $
in the left hand side of the given equation that is, ${\tan ^2}\theta {\cos ^2}\theta $ .
${\tan ^2}\theta {\cos ^2}\theta = \dfrac{{{{\sin }^2}\theta }}
{{{{\cos }^2}\theta }} \cdot {\cos ^2}\theta $
Cancel the numerator and denominator of the above expression that is, ${\cos ^2}\theta $
and ${\cos ^2}\theta $.
$\dfrac{{{{\sin }^2}\theta }}
{{{{\cos }^2}\theta }} \cdot {\cos ^2}\theta = {\sin ^2}\theta $
We know the trigonometric identity as, ${\sin ^2}\theta + {\cos ^2}\theta = 1$ .
We can rewrite the above identity in terms of ${\sin ^2}\theta $ as,
$
{\sin ^2}\theta + {\cos ^2}\theta = 1 \\
{\sin ^2}\theta = 1 - {\cos ^2}\theta \\
$
Substitute $1 - {\cos ^2}\theta $
for ${\sin ^2}\theta $
in the expression $\dfrac{{{{\sin }^2}\theta }}
{{{{\cos }^2}\theta }} \cdot {\cos ^2}\theta = {\sin ^2}\theta $.
$
\dfrac{{{{\sin }^2}\theta }}
{{{{\cos }^2}\theta }} \cdot {\cos ^2}\theta = {\sin ^2}\theta \\
= 1 - {\cos ^2}\theta \\
$
So we can write that ${\tan ^2}\theta {\cos ^2}\theta = 1 - {\cos ^2}\theta $ .
Thus, we get that the left hand side is equal to the right hand side that is, LHS = RHS.
Hence, the given statement is proved.
Note: To prove the given expression we must use the necessary trigonometric identities and ratios to simplify the expression. We can also prove the given expression by showing that the right hand side is equal to the left hand side that is, $1 - {\cos ^2}\theta = {\tan ^2}\theta {\cos ^2}\theta $
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