If cot θ = 7/8, evaluate (i) \large \frac{(1+sin\theta)(1-sin\theta)}{(1+cos\theta)(1-cos\theta)} (ii) \large cot^2\theta ?

byAmlesh Mahato •May 24, 2021

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Solution:

⇨ Given

⇨ cot θ = 7/8 = b/p

⇨ Therefore b = 7, p = 8, h = ?

⇨From Pythagoras Theorem

$If cot θ = 7/8, evaluate (i) \large \frac{(1+sin\theta)(1-sin\theta)}{(1+cos\theta)(1-cos\theta)} (ii) \large cot^2\theta ?$

$\large \Rightarrow h^{2}=8^{2}+7^{2}$

$\large \Rightarrow h^{2}=68+49$

$\large \Rightarrow h^{2}=113$

$\large \Rightarrow h=\sqrt{113}$

(i) $\large \frac{(1+sin\theta)(1-sin\theta)}{(1+cos\theta)(1-cos\theta)}$

$\large = \frac{1^{2}-sin^2\theta}{1^{2}-cos^2\theta}$

$\large = \frac{1-(\frac{p}{h})^2}{1-(\frac{b}{h})^2}$

$\large = \frac{1-(\frac{8}{\sqrt{113}})^2}{1-(\frac{7}{\sqrt{113}})^2}$

$\large = \frac{1-\frac{64}{113}}{1-\frac{49}{113}}$

$\large = \frac{\frac{113-64}{113}}{\frac{113-49}{113}}$

$\large = \frac{113-64}{113}\times \frac{113}{113-49}$

$\large = \frac{113-64}{113-49}$

$\large = \frac{49}{64}$

(ii) $\large cot^2\theta$

$\large =\left ( \frac{b}{p} \right )^{2}$

$\large =\left ( \frac{7}{8} \right )^{2}$

$\large =\frac{49}{64}$

Tags: Chapter-8 Exercise-8.1 Math- Class 10