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How do you prove $\dfrac{{1 - {{\tan }^2}\theta }}{{1 - {{\cot }^2}\theta }} = 1 - {\sec ^2}\theta$?

seo-qna
Last updated date: 23rd Feb 2024
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IVSAT 2024
Answer
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Hint: This question is from the topic of trigonometric identities. In this we need to prove $\dfrac{{1 - {{\tan }^2}\theta }}{{1 - {{\cot }^2}\theta }} = 1 - {\sec ^2}\theta $. To prove this we will use basic trigonometric identities and trigonometric functions. To prove this we start with L.H.S of the equation and write it in the form of $\sin \theta $ and $\cos \theta $.

Complete step by step solution:
Let us try to solve this question in which we are asked to prove
that $\dfrac{{1 - {{\tan }^2}\theta }}{{1 - {{\cot }^2}\theta }} = 1 - {\sec ^2}\theta $.
To prove this we will first use this relations $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and $\cot \theta = \dfrac{{\cos
\theta }}{{\sin \theta }}$. We simplify this equation to get the required result $1 - {\sec ^2}\theta$.
Let’s try to prove.
To Prove: $\dfrac{{1 - {{\tan }^2}\theta }}{{1 - {{\cot }^2}\theta }} = 1 - {\sec ^2}\theta $
Proof: We have,
$\dfrac{{1 - {{\tan }^2}\theta }}{{1 - {{\cot }^2}\theta }} = 1 - {\sec ^2}\theta $
$(1)$
Now by using the identities such as $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and $\cot \theta
= \dfrac{{\cos \theta }}{{\sin \theta }}$. Putting the values of these identities in equation$(1)$, we get
$\dfrac{{1 - {{\left( {\dfrac{{\sin \theta }}{{\cos \theta }}} \right)}^2}}}{{1 - {{\left(
{\dfrac{{\cos \theta }}{{\sin \theta }}} \right)}^2}}} = 1 - {\sec ^2}\theta $ $(2)$
$\dfrac{{1 - \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}}}{{1 - \dfrac{{{{\cos
}^2}\theta }}{{{{\sin }^2}\theta }}}} = 1 - {\sec ^2}\theta $ $(3)$
Now, by performing fraction subtraction in the L.H.S of equation $(3)$ numerator and denominator both, we get
$\dfrac{{\dfrac{{{{\cos }^2}\theta - {{\sin }^2}\theta }}{{{{\cos }^2}\theta
}}}}{{\dfrac{{{{\sin }^2}\theta - {{\cos }^2}\theta }}{{{{\sin }^2}\theta }}}} = 1 - {\sec ^2}\theta $
$(4)$
Now using this property $\dfrac{{\dfrac{a}{b}}}{{\dfrac{c}{d}}} = \dfrac{{a \cdot d}}{{b \cdot c}}$in the equation $(4)$, we get
$\dfrac{{{{\sin }^2}\theta ({{\cos }^2}\theta - {{\sin }^2}\theta )}}{{{{\cos
}^2}\theta ({{\sin }^2}\theta - {{\cos }^2}\theta )}} = 1 - {\sec ^2}\theta $ $(5)$
Now by using the result $\dfrac{{{{\cos }^2}\theta - {{\sin }^2}\theta }}{{{{\sin }^2}\theta - {{\cos }^2}\theta }} = - 1$ in the equation $(5)$, we get
$ - \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} = 1 - {\sec ^2}\theta $
$(6)$
As we already know that $\dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} = {\tan ^2}\theta $ in the equation $(6)$, we get
$ - {\tan ^2}\theta = 1 - {\sec ^2}\theta $ $(7)$
Now by using the trigonometric identity ${\sec ^2}\theta - {\tan ^2}\theta = 1$ in the equation $(7)$, we get
We write the above trigonometric identity as $ - {\tan ^2}\theta = 1 - {\sec ^2}\theta $. So we can write equation $(7)$ as,
$1 - {\sec ^2}\theta = 1 - {\sec ^2}\theta $
Since we have shown L.H.S of the equation $\dfrac{{1 - {{\tan }^2}\theta }}{{1 - {{\cot }^2}\theta }} = 1 - {\sec ^2}\theta $ equal to the R.H.S.
Hence proved.

Note: While solving these types of questions in which we have to prove trigonometric equations, we will start with the L.H.S of equation to derive R.H.S from it. To prove this question we only requires knowledge of basic trigonometric identities such as ${\sin ^2}x + {\cos ^2}x = 1$, ${\sec ^2}x - {\tan ^2}x = 1$