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Find $$\theta$$, if $$\displaystyle\frac{2\tan\displaystyle\frac{\theta}{2}}{1 + \tan^2\displaystyle\frac{\theta}{2}} = 1,\quad 0^{\small\circ} < \theta \le 90^{\small\circ}$$

Evaluate :$$\displaystyle \frac{5\sin ^{2}30^{\circ}+\cos ^{2}45^{\circ}+4\tan ^{2}60^{\circ}}{2\sin 30^{\circ}\cos 60^{\circ}+\tan 45^{\circ}}$$

$$2(\sin^{6}{\theta}+\cos^{6}{\theta})-3(\sin^{4}{\theta}+\cos^{4}{\theta})+1$$ is equal to

Value of $$\displaystyle\frac{\sin^{2}{{20}^{o}}+\cos^{4}{{20}^{o}}}{\sin^{4}{{20}^{o}}+\cos^{2}{{20}^{o}}}$$ is

If $$\displaystyle \tan \theta =\frac{x}{y},$$ then $$\displaystyle \frac{x\sin \theta +y\cos \theta }{x\sin \theta -y\cos \theta }$$ is equal to 

If $$\theta$$ is an acute angle such that $$\tan^{2}{\theta}=\dfrac{8}{7}$$,then the value of $$\displaystyle\frac{(1+\sin{\theta})(1-\sin{\theta})}{(1+\cos{\theta})(1-\cos{\theta})}$$ is

If $$\sin{x}+\sin^{2}{x}=1$$, then $$\cos^{12}{x}+3\cos^{10}{x}+3\cos^{8}{x}+\cos^{6}{x}-2$$ is equals to

If $$\displaystyle \tan 2A= \cot (A-60^{\circ}),$$ where 2A is an acute angle, then the value of A is 

If $$(\sec{A}-\tan{A})(\sec{B}-\tan{B})(\sec{C}-\tan{C})=(\sec{A}+\tan{A})(\sec{B}+\tan{B})(\sec{C}+\tan{C})$$ represents each side of a equilateral triangle, then each side is equal to -

If  $$\mathrm{cosec} {A}+\cot{A}=\cfrac{11}{2}$$, then $$\tan{A}$$ is equal to