Question 111
$$\frac{sec^{2}\theta-\cot^{2}(90^{\circ}-\theta)}{cosec^{2}67^{\circ}-\tan^{2}23^{\circ}}+sin^{2}40^{\circ}+sin^{2}50^{\circ}$$ is equal to
Solution
Expression : $$\frac{sec^{2}\theta-\cot^{2}(90^{\circ}-\theta)}{cosec^{2}67^{\circ}-\tan^{2}23^{\circ}}+sin^{2}40^{\circ}+sin^{2}50^{\circ}$$
= $$\frac{sec^2 \theta - tan^2 \theta}{cosec^2 67 - tan^2 (90 - 67)} + sin^2 40 + sin^2 (90 - 40)$$
= $$\frac{1}{cosec^2 67 - cot^2 67} + sin^2 40 + cos^2 40$$
= $$1 + 1 = 2$$
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