# If $cot\ θ = \frac{7}{8}$, evaluate:(i) $\frac{(1+sin\ θ)(1−sin\ θ)}{(1+cos\ θ)(1−cos\ θ)}$(ii) $cot^2\ θ$

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To do:

We have to evaluate:

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

(ii) $cot^2\ θ$

Solution:

(i) We know that,

In a right-angled triangle $ABC$ with a right angle at $B$,

By Pythagoras theorem,

$AC^2=AB^2+BC^2$

By trigonometric ratios definitions,

$sin\ A=\frac{Opposite}{Hypotenuse}=\frac{BC}{AC}$

$cos\ A=\frac{Adjacent}{Hypotenuse}=\frac{AB}{AC}$

Here,

Let $cot\ \theta=\frac{AB}{BC}=\frac{7}{8}$

$AC^2=AB^2+BC^2$

$\Rightarrow AC^2=(7)^2+(8)^2$

$\Rightarrow AC^2=49+64$

$\Rightarrow AC=\sqrt{113}$

Therefore,

$sin\ \theta=\frac{BC}{AC}=\frac{8}{\sqrt{113}}$

$cos\ \theta=\frac{AB}{AC}=\frac{7}{\sqrt{113}}$

$\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}=\frac{( 1+\frac{8}{\sqrt{113}})( 1-\frac{8}{\sqrt{113}})}{( 1+\frac{7}{\sqrt{113}})( 1-\frac{7}{\sqrt{113}})}$

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

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

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

$=\frac{49}{64}$

Therefore,  $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}=\frac{49}{64}$.

(ii) We know that,

$a^2=(a)^2$

Therefore,

$cot^2 \theta=(cot \theta)^2$

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

$=\frac{7^2}{8^2}$

$=\frac{49}{64}$

Hence, $cot^2 \theta=\frac{49}{64}$.

Updated on 10-Oct-2022 13:22:11