evaluate :- tan^2 60 +4 sin^2 45 + 3 sec^2 30 + 5 cos^2 90 / cosec 30 + sec 60 - cot^2 30

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Solution

$\frac{\tan^{2} 60 ° + 4 \sin^{2} 45 ° + 3 \sec^{2} 30 ° + 5 \cos^{2} 90 °}{cosec 30 ° + \sec 60 ° - \cot^{2} 30 °} We know that \tan 60 ° = \sqrt{3}; \sin 45 ° = \frac{1}{\sqrt{2}}; \sec 30 ° = \frac{2}{\sqrt{3}}; \cos 90 ° = 0; cosec 30 ° = 2; \sec 60 ° = 2; \cot 30 ° = \sqrt{3} \frac{\tan^{2} 60 ° + 4 \sin^{2} 45 ° + 3 \sec^{2} 30 ° + 5 \cos^{2} 90 °}{cosec 30 ° + \sec 60 ° - \cot^{2} 30 °} = \frac{3 + 4 \times \frac{1}{2} + 3 \times \frac{4}{3} + 0}{2 + 2 - 3} = \frac{3 + 2 + 4}{1} = 9$

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{tan}^{2} 30 \cdot sin 30 + cos 60 \cdot {sin}^{2} 90 \cdot {tan}^{2} 60 - 2 tan 45 \cdot {cos}^{2} 0 \cdot sin 90

s i n^{2} 30^{\circ} + c o s^{2} 60 + {tan}^{2} 45^{\circ} + s e c^{2} 60^{\circ} - c o {sec}^{2} 30^{\circ} = \frac{3}{2}

2 t a n^{2} 45^{\circ} + 3 c o s^{2} 30^{\circ} - s i n^{2} 60^{\circ}

\frac{2 {cos}^{2} 60^{\circ} + 3 {sec}^{2} 30^{\circ} - 2 {tan}^{2} 45^{\circ}}{{sin}^{2} 30^{\circ} + {cos}^{2} 45^{\circ}}

2 ({cos}^{2} 45^{\circ} + {tan}^{2} 60^{\circ}) - 6 ({sin}^{2} 45^{\circ} - {tan}^{2} 30^{\circ}) = 6

2 ({cos}^{4} 60^{\circ} + {sin}^{4} 30^{\circ}) - ({tan}^{2} 60^{\circ} + {cos}^{2} 45^{\circ}) + 3 {sec}^{2} 30^{\circ} = \frac{3}{4}

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