Question

Prove that, \[\dfrac{1}{\sec A+\tan A}-\dfrac{1}{\cos A}=\dfrac{1}{\cos A}-\dfrac{1}{\sec A-\tan A}\]

Answer 1

Hint: We can solve this question by using trigonometric identity ${{\sec }^{2}}A-{{\tan }^{2}}A=1$.

Complete step-by-step answer:
Given expression is \[\dfrac{1}{\sec A+\tan A}-\dfrac{1}{\cos A}=\dfrac{1}{\cos A}-\dfrac{1}{\sec A-\tan A}\]
On taking L.H.S
\[\dfrac{1}{\sec A+\tan A}-\dfrac{1}{\cos A}\]
We can multiply numerator and denominator by $\sec A-\tan A$ in first term
\[\Rightarrow \dfrac{1}{\sec A+\tan A}\times \dfrac{\sec A-\tan A}{(\sec A-\tan A)}-\sec A\] $\left\{ \because \dfrac{1}{\cos A}=\sec A \right\}$
\[\Rightarrow \dfrac{\sec A-\tan A}{({{\sec }^{2}}A-{{\tan }^{2}}A)}-\sec A\]
\[\Rightarrow \dfrac{\sec A-\tan A}{1}-\sec A\] $\left\{ \because {{\sec }^{2}}A-{{\tan }^{2}}A=1 \right\}$
\[\Rightarrow \sec A-\tan A-\sec A\]
\[\Rightarrow -\tan A\]

On taking R.H.S
\[\dfrac{1}{\cos A}-\dfrac{1}{\sec A-\tan A}\]
We can multiply numerator and denominator by $\sec A+\tan A$ in second term
\[\Rightarrow \sec A-\dfrac{1}{\sec A-\tan A}\times \dfrac{\sec A+\tan A}{(\sec A+\tan A)}\] $\left\{ \because \dfrac{1}{\cos A}=\sec A \right\}$
\[\Rightarrow \sec A-\dfrac{(\sec A+\tan A)}{({{\sec }^{2}}A-{{\tan }^{2}}A)}\]
\[\Rightarrow \sec A-\sec A-\tan A\] $\left\{ \because {{\sec }^{2}}A-{{\tan }^{2}}A=1 \right\}$
\[\Rightarrow -\tan A\]
On simplifying L.H.S equal to R.H.S.

Note: We can also do this question by converting expression in terms of sinA and cosA.
As we have $\tan A=\dfrac{\sin A}{\cos A},\sec A=\dfrac{1}{\cos A}$