Name: Trigonometric Ratios of Complementary Angles
Uploaded: 2019-07-19
Duration: 9 min 3 s
Description: What is the connection between the trigonometric ratios of an angle and its complementary angle? This video not only gives you that but helps you derive them...

Question

\frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A} = \frac{2}{1 - 2 \cos^{2} A} = \frac{2 \sec^{2} A}{\tan^{2} A - 1}

Accepted Answer

Given, $\frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A} = \frac{2}{1 - 2 \cos^{2} A} = \frac{2 \sec^{2} A}{\tan^{2} A - 1}$

Let's take first, $\frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A}$

$\Rightarrow \frac{{(\sin A + \cos A)}^{2} + {(\sin A - \cos A)}^{2}}{(\sin A - \cos A) (\sin A + \cos A)}$

Applying the identity, ${(a + b)}^{2} = a^{2} + b^{2} + 2 a b$

$\Rightarrow \frac{\sin^{2} A + \cos^{2} A + 2 \sin A \cos A + \sin^{2} A + \cos^{2} A - 2 \sin A \cos A}{\sin^{2} A - \cos^{2} A}$

$\Rightarrow \frac{2 (\sin^{2} A + \cos^{2} A)}{\sin^{2} A - \cos^{2} A}$

We know that, $\sin^{2} A + \cos^{2} A = 1$

$\Rightarrow \frac{2}{1 - \cos^{2} A - \cos^{2} A} = \frac{2}{1 - 2 \cos^{2} A}$

Thus, $\frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A} = \frac{2}{1 - 2 \cos^{2} A} . . . (1)$

Nw, let's take $\frac{2 \sec^{2} A}{\tan^{2} A - 1}$

Again, $\sec A = \frac{1}{\cos A}$ and $\tan A = \frac{\sin A}{\cos A}$

Thus, $\frac{2 \sec^{2} A}{\tan^{2} A - 1} = \frac{\frac{2}{\cos^{2} A}}{\frac{\sin^{2} A}{\cos^{2} A} - 1}$

$\Rightarrow \frac{\frac{2}{\cos^{2} A}}{\frac{\sin^{2} A - \cos^{2} A}{\cos^{2} A}} = \frac{2}{\cos^{2} A} \times \frac{\cos^{2} A}{\sin^{2} A - \cos^{2} A}$

$\Rightarrow \frac{2}{1 - \cos^{2} A - \cos^{2} A} = \frac{2}{1 - 2 \cos^{2} A}$

$\frac{2 \sec^{2} A}{\tan^{2} A - 1} = \frac{2}{1 - 2 \cos^{2} A} . . . (2)$

Therefore, from the equations $(1)$ and $(2)$ , we get,

$\frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A} = \frac{2}{1 - 2 \cos^{2} A} = \frac{2 \sec^{2} A}{\tan^{2} A - 1}$

Hence proved.

$\frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A} = \frac{2}{1 - 2 \cos^{2} A} = \frac{2 \sec^{2} A}{\tan^{2} A - 1}$

Important Questions on Trigonometric Identities

Learn and Prepare for any exam you want !

sin A+cos Asin A-cos A+sin A-cos Asin A+cos A=21-2cos2 A=2 sec2 Atan2 A-1

Important Questions on Trigonometric Identities

Learn and Prepare for any exam you want !

$\frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A} = \frac{2}{1 - 2 \cos^{2} A} = \frac{2 \sec^{2} A}{\tan^{2} A - 1}$