સાબિત કરો કે,frac{sin theta}{1+cos theta}+fra | vedclass

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(N/A) ડા.બા. $= \frac{\sin 	heta}{1+\cos 	heta} + \frac{1+\cos 	heta}{\sin 	heta}$$= \frac{\sin^2 	heta + (1+\cos 	heta)^2}{\sin 	heta(1+\cos \

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(N/A) ડા.બા. $= \frac{\sin 	heta}{1+\cos 	heta} + \frac{1+\cos 	heta}{\sin 	heta}$$= \frac{\sin^2 	heta + (1+\cos 	heta)^2}{\sin 	heta(1+\cos 	heta)}$$= \frac{\sin^2 	heta + 1 + \cos^2 	heta + 2 \cos 	heta}{\sin 	heta(1+\cos 	heta)}$ $\quad [\because (a+b)^2 = a^2 + b^2 + 2ab]$$= \frac{(\sin^2 	heta + \cos^2 	heta) + 1 + 2 \cos 	heta}{\sin 	heta(1+\cos 	heta)}$$= \frac{1 + 1 + 2 \cos 	heta}{\sin 	heta(1+\cos 	heta)}$ $\quad [\because \sin^2 	heta + \cos^2 	heta = 1]$$= \frac{2 + 2 \cos 	heta}{\sin 	heta(1+\cos 	heta)}$$= \frac{2(1+\cos 	heta)}{\sin 	heta(1+\cos 	heta)}$$= \frac{2}{\sin 	heta} = 2 \operatorname{cosec} 	heta = 	ext{જ.બા.}$ $\quad [\because \operatorname{cosec} 	heta = \frac{1}{\sin 	heta}]$

સાબિત કરો કે,
$\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}=2 \operatorname{cosec} \theta$

Similar Questions

સાબિત કરો કે:
$(i)$ $\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ} = 1$
$(ii)$ $\cos 38^{\circ} \cos 52^{\circ} - \sin 38^{\circ} \sin 52^{\circ} = 0$

$\frac{\cos ^{2} 40^{\circ}+\cos ^{2} 50^{\circ}}{\sin ^{2} 40^{\circ}+\sin ^{2} 50^{\circ}}=\ldots \ldots \ldots \ldots$

જો $\sec \theta = \frac{5}{3}$ હોય,તો $\tan \theta = \ldots$

$\sec 55^{\circ} \cdot \sin 35^{\circ} + \cos 35^{\circ} \cdot \operatorname{cosec} 55^{\circ} = \ldots \ldots \ldots \ldots$

જો $1+\sin ^{2} \theta=3 \sin \theta \cos \theta$ હોય,તો સાબિત કરો કે $\tan \theta=1$ અથવા $\frac{1}{2}$ થાય.

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સાબિત કરો કે,$\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}=2 \operatorname{cosec} \theta$