Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# If Sin X = a 2 − B 2 a 2 + B 2 , Then the Values of Tan X, Sec X and Cosec X - Mathematics

If $\sin x = \frac{a^2 - b^2}{a^2 + b^2}$, then the values of tan x, sec x and cosec x

#### Solution

$\sin x = \frac{a^2 - b^2}{a^2 + b^2}$
We know:
$\sin^2 x + \cos^2 x = 1$
$\cos^2 x = 1 - \sin^2 x$
$= 1 - \left( \frac{a^2 - b^2}{a^2 + b^2} \right)^2$
$= \frac{\left( a^4 + b^4 + 2 a^2 b^2 \right) - \left( a^4 + b^4 - 2 a^2 b^2 \right)}{\left( a^2 + b^2 \right)^2}$
$= \frac{4 a^2 b^2}{\left( a^2 + b^2 \right)^2}$
$\Rightarrow \cos x = \frac{2ab}{\left( a^2 + b^2 \right)}$
$\tan x = \frac{\sin x}{\cos x} = \frac{\frac{a^2 - b^2}{a^2 + b^2}}{\frac{2ab}{a^2 + b^2}} = \frac{a^2 - b^2}{2ab}$
$\sec x = \frac{1}{\cos x} = \frac{a^2 + b^2}{2ab}$
$cosec x = \frac{1}{\sin x} = \frac{a^2 + b^2}{a^2 - b^2}$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 5 Trigonometric Functions
Exercise 5.1 | Q 18 | Page 18