કિંમત શોધો :
$\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}$
$\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}$
$=\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2}=\frac{\frac{1}{\sqrt{2}}}{\frac{2+2 \sqrt{3}}{\sqrt{3}}}$
$=\frac{\sqrt{3}}{\sqrt{2}(2+2 \sqrt{3})}=\frac{\sqrt{3}}{2 \sqrt{2}+2 \sqrt{6}}$
$=\frac{\sqrt{3}(2 \sqrt{6}-2 \sqrt{2})}{(2 \sqrt{6}+2 \sqrt{2})(2 \sqrt{6}-2 \sqrt{2})}$
$=\frac{2 \sqrt{3}(\sqrt{6}-\sqrt{2})}{(2 \sqrt{6})^{2}-(2 \sqrt{2})^{2}}=\frac{2 \sqrt{3}(\sqrt{6}-\sqrt{2})}{24-8}=\frac{2 \sqrt{3}(\sqrt{6}-\sqrt{2})}{16}$
$=\frac{\sqrt{18}-\sqrt{6}}{8}=\frac{3 \sqrt{2}-\sqrt{6}}{8}$
$(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)=.......$
કિંમત શોધો : $\frac{\tan 65^{\circ}}{\cot 25^{\circ}}$
કિંમત શોધો :
$\frac{\sin 30^{\circ}+\tan 45^{\circ}-\operatorname{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}$
કિંમત શોધો :
$\cos 48^{\circ}-\sin 42^{\circ}$
લધુ કોણ $\angle B$ તથા $\angle Q$ માટે $\sin B =\sin Q$ છે. સાબિત કરો કે $\angle B =\angle Q$.