Question : The value of $\frac{3\left(\cot ^2 47^{\circ}-\sec ^2 43^{\circ}\right)-2\left(\tan ^2 23^{\circ}-\operatorname{cosec}^2 67^{\circ}\right)}{\operatorname{cosec}^2\left(68^{\circ}+\theta\right)-\tan \left(\theta+61^{\circ}\right)-\tan ^2\left(22^{\circ}-\theta\right)+\cot \left(29^{\circ}-\theta\right)}$ is:
Option 1: –1
Option 2: 1
Option 3: 5
Option 4: 0
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Correct Answer: –1
Solution : Given: $\frac{3\left(\cot ^2 47^{\circ}-\sec ^2 43^{\circ}\right)-2\left(\tan ^2 23^{\circ}-\operatorname{cosec}^2 67^{\circ}\right)}{\operatorname{cosec}^2\left(68^{\circ}+\theta\right)-\tan \left(\theta+61^{\circ}\right)-\tan ^2\left(22^{\circ}-\theta\right)+\cot \left(29^{\circ}-\theta\right)}$ = $\frac{3\left(\cot ^2(90^{\circ}- 43^{\circ})-\sec ^2 43^{\circ}\right)-2\left(\tan ^2 23^{\circ}-\operatorname {cosec^2} (90^{\circ}-23^{\circ})\right)}{\sec^2\left(90^{\circ}-(68^{\circ}+\theta)\right)-\cot \left(90^{\circ}-(\theta+61^{\circ})\right)-\tan ^2\left(22^{\circ}-\theta\right)+\cot \left(29^{\circ}-\theta\right)}$ = $\frac{3\left(\tan ^243^{\circ}-\sec ^2 43^{\circ}\right)-2\left(\tan ^2 23^{\circ}-\sec^2 23^{\circ}\right)}{\sec^2\left(22^{\circ}-\theta\right)-\cot \left(29^{\circ}-\theta\right)-\tan ^2\left(22^{\circ}-\theta\right)+\cot \left(29^{\circ}-\theta\right)}$ = $\frac{(3× -1)-(2×(-1))}{1}$ = $\frac{-3+2}{1}$ = $-1$ Hence, the correct answer is –1.
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Question : The value of
Question : $\left(\frac{\tan ^3 \theta}{\sec ^2 \theta}+\frac{\cot ^3 \theta}{\operatorname{cosec}^2 \theta}+2 \sin \theta \cos \theta\right) \div\left(1+\operatorname{cosec}^2 \theta+\tan ^2 \theta\right), 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : Let $0^{\circ}<\theta<90^{\circ}$, $\left(1+\cot ^2 \theta\right)\left(1+\tan ^2 \theta\right) × (\sin \theta-\operatorname{cosec} \theta)(\cos \theta-\sec \theta)$ is equal to:
Question : $\frac{(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)}{(\sec \theta+\tan \theta)(1-\sin \theta)}$ is equal to:
Question : The expression $(\tan \theta+\cot \theta)(\sec \theta+\tan \theta)(1-\sin \theta), 0^{\circ}<\theta<90^{\circ}$, is equal to:
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