Question

If $cot\theta =\frac{7}{8}$, evaluate:
$\frac{(1+sin\theta )(1-sin\theta )}{(1+cos\theta )(1-cos\theta )}$

• A. $\frac{49}{64}$
• B. $\frac{49}{8}$
• C. $\frac{7}{64}$
• D. $\frac{7}{8}$

Answer 1

The correct option is A $\frac{49}{64}$

Let $\Delta ABC$ in which $\angle B={90}^{\circ }$ and $\angle C=\theta$
According to question
$cot\theta =\frac{BC}{AB}=\frac{7}{8}$
Let BC = 7k and AB = 8k, where k is a positive real number.
By Pythagoras theorem in $\Delta ABC$ we get
$A{C}^2=A{B}^2+B{C}^2$
$A{C}^2=(8k{)}^2+(7k{)}^2$
$A{C}^2=64k^2+49k^2$
$A{C}^2=113k^2$
$AC=\sqrt{113}k$
$sin\theta =\frac{AB}{AC}=\frac{8k}{\sqrt{113}k}=\frac{8}{\sqrt{113}}$
and $cos\theta =\frac{BC}{AC}=\frac{7k}{\sqrt{113}k}=\frac{7}{\sqrt{113}}\dots \dots \left(i\right)$
$\frac{(1+sin\theta )(1-sin\theta )}{(1+cos\theta )(1-cos\theta )}=\frac{(1-si{n}^2\theta )}{(1-co{s}^2\theta )}=\frac{\{1-{\left(\frac{8}{\sqrt{113}}\right)}^2\}}{\{1-{\left(\frac{7}{\sqrt{113}}\right)}^2\}}=\frac{\{1-\left(\frac{64}{113}\right)\}}{\{1-\left(\frac{49}{113}\right)\}}=\frac{\left\{\frac{(113-64)}{113}\right\}}{\left\{\frac{(113-49)}{113}\right\}}=\frac{49}{64}$