Question

If $cot\Theta =\frac{7}{8}$, evaluate :
(i) ($\frac{(1+sin\Theta)(1-sin\Theta)}{(1+cos\Theta)(1-cos\Theta)}$
(ii) cot² θ

Answer 1

SOLUTION IS IN THE ATTACHMENT.

** Trigonometry is the study of the relationship between the sides and angles of a triangle.

The ratio of the sides of a right angled triangle with respect to its acute angles are called trigonometric ratios.

** For any acute angle in a right angle triangle the side opposite to the acute angle is called a perpendicular(P),  the side adjacent to this acute angle is called the base(B) and side opposite to the right angle is called the hypotenuse(H).

** Find the third  side of the right ∆ ABC by using Pythagoras theorem (AC² = AB² + BC²).

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

Solution :

Consider a Δ ABC in which ∠A = Θ° and ∠B = 90°.

Then, Base = AB, Perpendicular = BC and Hypotenuse = AC

∴ cot Θ = $\frac{Base}{Perp} = \frac{AB}{BC} = \frac{7}{8}$

Let AB =7k and BC = 8k.

Then, AC = $\sqrt{BC^{2} + AB^{2}}$ = $\sqrt{(8k)^{2} +(7k)^{2}}$

               = $\sqrt{64k^{2} + 49k^{2}}$ = $\sqrt{113k^{2}} = \sqrt{113k}$

∴ sin Θ = $\frac{Perp}{Hyp} = \frac{BC}{AC} = \frac{8k}{\sqrt{113k}} = \frac{8}{\sqrt{113}}$

and, cos Θ = $\frac{Base}{Hyp} = \frac{AB}{AC} = \frac{7k}{\sqrt{113k}} = \frac{7}{\sqrt{113}}$

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

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

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

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

(ii) $cot^{2} \ Θ = (\frac{7}{8})^{2} = \frac{49}{64}$