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If $cot\ θ = \frac{7}{8}$, evaluate:
(i) $\frac{(1+sin\ θ)(1−sin\ θ)}{(1+cos\ θ)(1−cos\ θ)}$
(ii) $cot^2\ θ$
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, $cot^2 \theta=\frac{49}{64}$.
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