InterviewSolution
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InterviewSolution
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If tanθ = 7/8, then the value of (1+sin θ)(1−sin θ)/(1+cos θ)(1−cos θ) ……..A) 64/49B) 49/64C) 8/7D) 7/8 |
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Answer» Correct option is: A) \(\frac{64}{49}\) We have tan\(\theta\) = \(\frac{7}{8}\) Now, \(\frac{(1+sin \theta)(1-sin \theta)}{1+cos \theta)(1-cos \theta)}\) = \(\frac {1-sin^2\theta}{1-cos^2\theta} = \frac {cos^2\theta}{sin^2\theta} \) (\(\because\) \(sin^2\theta + cos^2\theta = 1\)) = \((\frac {cos\theta}{sin\theta})^2 = cot^2\theta\) = \(\frac {1}{tan^2 \theta} = \frac {1}{(\frac 78)^2} = \frac {8^2}{7^2}\) = \(\frac {64}{49}\) Correct option is: A) \(\frac{64}{49}\) |
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