
Prove that ${\tan ^2}\theta {\cos ^2}\theta = 1 - {\cos ^2}\theta $
Answer
554.1k+ 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 $
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

10 examples of friction in our daily life

