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