Question

If 5 sin Theta equal to 4 prove that 1 bys cos theta + 1 by cot theta equal to 3

Answer 1

Answer:

In the attachment.. Hope it's helpful..

Answer 2

Given that :

$5 \sin\alpha = 4 \\ \\ = > \sin\alpha = \frac{4}{5}$

As we know that :

$\sin\alpha = \frac{Perpendicular}{Hypotenuse}$

Then,

$Perpendicular \: = 4 \: and \: Hypotenuse = 5$

According to the Pythagoras theoram :

${(Hypotenuse)}^{2} = {(Perpendicular)}^{2} + {(Base)}^{2} \\ \\ = > {(Base)}^{2} = {5}^{2} - {4}^{2} \\ \\ = > {(Base)}^{2} = 25 - 16 = 9 \\ \\ = > Base = 3$

Now,

$\cos\alpha = \frac{Base}{Hypotenuse} = \frac{3}{5} \: and \: \cot \alpha = \frac{Base}{Perpendicular} = \frac{3}{4}$

We have to prove that :

$\frac{1}{ \cos\alpha } + \frac{1}{ \cot \alpha } = 3$

On taking LHS :

$\frac{1}{ \cos\alpha } + \frac{1}{ \cot\alpha } \\ \\ = > \frac{1}{ \frac{3}{5} } + \frac{1}{ \frac{3}{4} } \\ \\ = > \frac{5}{3} + \frac{4}{3} \\ \\ = > \frac{9}{3} \\ \\ = > 3 = RHS \\ \\ HENCE \: PROVED$