(B) Given expression: $\frac{\sin x}{1+\cos x} + \frac{1+\cos x}{\sin x}$ Taking the common denominator: $\frac{\sin^2 x + (1+\cos x)^2}{\sin x(1+\cos

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(B) Given expression: $\frac{\sin x}{1+\cos x} + \frac{1+\cos x}{\sin x}$ Taking the common denominator: $\frac{\sin^2 x + (1+\cos x)^2}{\sin x(1+\cos x)}$ $= \frac{\sin^2 x + 1 + 2\cos x + \cos^2 x}{\sin x(1+\cos x)}$ Since $\sin^2 x + \cos^2 x = 1$,we get: $\frac{1 + 1 + 2\cos x}{\sin x(1+\cos x)}$ $= \frac{2 + 2\cos x}{\sin x(1+\cos x)} = \frac{2(1+\cos x)}{\sin x(1+\cos x)}$ $= \frac{2}{\sin x} = 2 \operatorname{cosec} x$

Class 11 Mathematics Trigonometrical Ratios, Functions and Identities Fundamental trigonometrical ratios and functions, Trigonometrical ratio of allied angles

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$\frac{\sin x}{1+\cos x} + \frac{1+\cos x}{\sin x} = ?$

AP EAMCET 2022 All AP EAMCET

A
$2 \sec x$
B
$2 \operatorname{cosec} x$
C
$\tan 2x$
D
$\sin 2x$

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Class 11

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Trigonometrical Ratios, Functions and Identities

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Fundamental trigonometrical ratios and functions, Trigonometrical ratio of allied angles

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$\frac{\sin x}{1+\cos x} + \frac{1+\cos x}{\sin x} = ?$

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$\frac{{\cot^2 15^\circ - 1}}{{\cot^2 15^\circ + 1}} = $

If $\alpha$ lies in the second quadrant,then $\sqrt{\frac{1 - \sin \alpha}{1 + \sin \alpha}} - \sqrt{\frac{1 + \sin \alpha}{1 - \sin \alpha}} = $

Find the angle in radian through which a pendulum swings if its length is $75\, cm$ and the tip describes an arc of length $15\, cm$.

If $\cot \theta = -\frac{2}{3}$ and $\theta$ does not lie in the $4^{\text{th}}$ quadrant,then $\frac{(5 \sin \theta + \cos \theta)^2}{\tan \theta + \cot \theta} = $

Find the radian measure corresponding to the following degree measure: $520^{\circ}$

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