Prove that cos 0- -sin-2-cos2-6-2-4tan-3 -3-3-38

→cos0 ^ 0 .sinπ #160; 2 .cos^ 2 π #160; 6 ^ 2 π #160; 4 #160; tanπ #160; 3 +π #160; 3 #160; → 1× 1× 3 4 × 2 √( 3 ) + 1 √( 3 ) ...

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Sign of Trigonometric Ratios in Different Quadrants

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Question

Prove that $\frac{cos 0^{\circ} . sin \frac{π}{2} . {cos}^{2} \frac{π}{6} . {sec}^{2} \frac{π}{4}}{tan \frac{π}{3} + cot \frac{π}{3}} = \frac{3 \sqrt{3}}{8}$

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\frac{cos 0 \cdot sin \frac{π}{2} \cdot {cos}^{2} (\frac{π}{6}) \cdot {sec}^{2} (\frac{π}{4})}{tan \frac{π}{3} + cot \frac{π}{3}}

sin \frac{π}{6} cos 0 + sin \frac{π}{4} cos \frac{π}{4} + sin \frac{π}{3} cos \frac{π}{6} = \frac{7}{4}

{sin}^{2} \frac{π}{6} + {cos}^{2} \frac{π}{3} - {tan}^{2} \frac{π}{4} = - \frac{1}{2}

{sin}^{2} \frac{π}{6} + {cos}^{2} \frac{π}{3} - {tan}^{2} \frac{π}{4} = - \frac{1}{2}

{sin}^{2} \frac{π}{6} + {cos}^{2} \frac{π}{3} - {tan}^{2} \frac{π}{4} = - \frac{1}{2}

tan (\frac{π}{4} + \frac{y}{2})

{tan}^{3} (\frac{π}{4} + \frac{x}{2})

sin y = sin x . \frac{3 + {sin}^{2} x}{1 + 3 {sin}^{2} x} .

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