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Question: Prove that \({\tan ^2}\theta {\cos ^2}\theta = 1 - {\cos ^2}\theta \)...

Prove that tan2θcos2θ=1cos2θ{\tan ^2}\theta {\cos ^2}\theta = 1 - {\cos ^2}\theta

Explanation

Solution

In order to prove the given equation, we need to use the trigonometric ratio tanθ=sinθcosθ\tan \theta = \dfrac{{\sin \theta }} {{\cos \theta }}. Then, we need to use the trigonometric identity sin2θ=1cos2θ{\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 tan2θcos2θ=1cos2θ{\tan ^2}\theta {\cos ^2}\theta = 1 - {\cos ^2}\theta .
In the above equation, the left hand side is tan2θcos2θ{\tan ^2}\theta {\cos ^2}\theta
and the right hand side is 1cos2θ1 - {\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 tanθ\tan \theta
is, tanθ=sinθcosθ\tan \theta = \dfrac{{\sin \theta }} {{\cos \theta }}.
Substitute sinθcosθ\dfrac{{\sin \theta }} {{\cos \theta }}
for tanθ\tan \theta
in the left hand side of the given equation that is, tan2θcos2θ{\tan ^2}\theta {\cos ^2}\theta .
tan2θcos2θ=sin2θcos2θcos2θ{\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, cos2θ{\cos ^2}\theta
and cos2θ{\cos ^2}\theta .
sin2θcos2θcos2θ=sin2θ\dfrac{{{{\sin }^2}\theta }} {{{{\cos }^2}\theta }} \cdot {\cos ^2}\theta = {\sin ^2}\theta
We know the trigonometric identity as, sin2θ+cos2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 .
We can rewrite the above identity in terms of sin2θ{\sin ^2}\theta as,
sin2θ+cos2θ=1 sin2θ=1cos2θ  {\sin ^2}\theta + {\cos ^2}\theta = 1 \\\ {\sin ^2}\theta = 1 - {\cos ^2}\theta \\\
Substitute 1cos2θ1 - {\cos ^2}\theta
for sin2θ{\sin ^2}\theta
in the expression sin2θcos2θcos2θ=sin2θ\dfrac{{{{\sin }^2}\theta }} {{{{\cos }^2}\theta }} \cdot {\cos ^2}\theta = {\sin ^2}\theta .
sin2θcos2θcos2θ=sin2θ =1cos2θ  \dfrac{{{{\sin }^2}\theta }} {{{{\cos }^2}\theta }} \cdot {\cos ^2}\theta = {\sin ^2}\theta \\\ = 1 - {\cos ^2}\theta \\\
So we can write that tan2θcos2θ=1cos2θ{\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, 1cos2θ=tan2θcos2θ1 - {\cos ^2}\theta = {\tan ^2}\theta {\cos ^2}\theta