Question

If $\cot \theta =\frac{7}{8},$ evaluate :

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

(ii) ${\cot }^2\theta$

Answer 1

Given,

$cot\theta =\frac{7}{8}$

$\tan \theta =\frac{1}{\cot \theta }=\frac{8}{7}$

We know that,

$\tan \theta =\frac{oppositeSide}{adjacentSide}$

From Pythagoras theorem,

$Hypotenus{e}^2=OppositeSid{e}^2+AdjacentSid{e}^2$

$Hypotenus{e}^2=8^2+7^2$

$Hypotenus{e}^2=64+49=113$

$Hypotenuse=\sqrt{113}$

$\sin \theta =\frac{oppositeSide}{Hypotenuse}=\frac{8}{\sqrt{113}}$

$\cos \theta =\frac{AdjacentSide}{Hypotenuse}=\frac{7}{\sqrt{113}}$

Solution(i):

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

We have, $a^2-b^2=(a+b)(a-b)$

Similarly,

$(1-{\sin }^2\theta )=(1+\sin \theta )(1-\sin \theta )$

$(1-{\cos }^2\theta )=(1+\cos \theta )(1-\cos \theta )$

Therefore,

$\frac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}=\frac{(1-{\sin }^2\theta )}{(1-{\cos }^2\theta )}$

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

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

Solution(ii):

Given,

$cot\theta =\frac{7}{8}$

$co{t}^2\theta =(\frac{7}{8}{)}^2=\frac{49}{64}$