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8. Introduction to Trigonometry
medium
Evaluate the following:
$\frac{\sin 30^{\circ}+\tan 45^{\circ}-\operatorname{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}$
Option A
Option B
Option C
Option D
Solution
$\frac{\sin 30^{\circ}+\tan 45^{\circ}-\operatorname{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}$
$=\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1}=\frac{\frac{3}{2}-\frac{2}{\sqrt{3}}}{\frac{3}{2}+\frac{2}{\sqrt{3}}}$
$=\frac{\frac{3 \sqrt{3}-4}{2 \sqrt{3}}}{\frac{3 \sqrt{3}+4}{2 \sqrt{3}}}=\frac{(3 \sqrt{3}-4)}{(3 \sqrt{3}+4)}$
$=\frac{(3 \sqrt{3}-4)(3 \sqrt{3}-4)}{(3 \sqrt{3}+4)(3 \sqrt{3}-4)}=\frac{(3 \sqrt{3}-4)^{2}}{(3 \sqrt{3})^{2}-(4)^{2}}$
$=\frac{27+16-24 \sqrt{3}}{27-16}=\frac{43-24 \sqrt{3}}{11}$
Standard 10
Mathematics
