Question

if 3cotA =4,then find the value of 3cosA+2sinA
__________
3cosA-2sinA​

Answer 1

Answer:

$\displaystyle{\boxed{\red{\sf\:\dfrac{3\:\cos\:A\:+\:2\:\sin\:A}{3\:\cos\:A\:-\:2\:\sin\:A}\:=\:3\:}}}$

Step-by-step-explanation:

We have given that,

$\displaystyle{\sf\:3\:\cot\:A\:=\:4}$

We have to find the value of

$\displaystyle{\sf\:\dfrac{3\:\cos\:A\:+\:2\:\sin\:A}{3\:\cos\:A\:-\:2\:\sin\:A}}$

Now,

$\displaystyle{\sf\:3\:\cot\:A\:=\:4}$

$\displaystyle{\implies\boxed{\sf\:\cot\:A\:=\:\dfrac{4}{3}}}$

Consider a right triangle ABC right angled at angle B.

We know that,

$\displaystyle{\pink{\sf\:\cot\:A\:=\:\dfrac{Adjacent\:side}{Opposite\:side}}}$

$\displaystyle{\implies\sf\:\dfrac{4}{3}\:=\:\dfrac{AB}{BC}}$

$\displaystyle{\implies\sf\:AB\:=\:4x}$

$\displaystyle{\sf\:BC\:=\:3x}$

Now, by Pythagoras theorem,

$\displaystyle{\pink{\sf\:(\:AC\:)^2\:=\:(\:AB\:)^2\:+\:(\:BC\:)^2}}$

$\displaystyle{\implies\sf\:AC^2\:=\:(\:4x\:)^2\:+\:(\:3x\:)^2}$

$\displaystyle{\implies\sf\:AC^2\:=\:16x^2\:+\:9x^2}$

$\displaystyle{\implies\sf\:AC^2\:=\:25x^2}$

$\displaystyle{\implies\sf\:AC\:=\:5x\:}$

$\displaystyle{\therefore\boxed{\sf\:Hypotenuse\:=\:5x}}$

Now, we know that,

$\displaystyle{\pink{\sf\:\sin\:A\:=\:\dfrac{Opposite\:side}{Hypotenuse}}}$

$\displaystyle{\implies\sf\:\sin\:A\:=\:\dfrac{3\:\cancel{x}}{5\:\cancel{x}}}$

$\displaystyle{\implies\sf\:\sin\:A\:=\:\dfrac{3}{5}}$

$\displaystyle{\implies\sf\:2\:\sin\:A\:=\:\dfrac{2\:\times\:3}{5}}$

$\displaystyle{\implies\boxed{\blue{\sf\:2\:\sin\:A\:=\:\dfrac{6}{5}}}}$

Now, we know that,

$\displaystyle{\pink{\sf\:\cos\:A\:=\:\dfrac{Adjacent\:side}{Hypotenuse}}}$

$\displaystyle{\implies\sf\:\cos\:A\:=\:\dfrac{4\:\cancel{x}}{5\:\cancel{x}}}$

$\displaystyle{\implies\sf\:\cos\:A\:=\:\dfrac{4}{5}}$

$\displaystyle{\implies\sf\:3\:\cos\:A\:=\:\dfrac{3\:\times\:4}{5}}$

$\displaystyle{\implies\boxed{\green{\sf\:3\:\cos\:A\:=\:\dfrac{12}{5}}}}$

Now, we have to find the value of

$\displaystyle{\sf\:\dfrac{3\:\cos\:A\:+\:2\:\sin\:A}{3\:\cos\:A\:-\:2\:\sin\:A}}$

$\displaystyle{\implies\sf\:\dfrac{\dfrac{12}{5}\:+\:\dfrac{6}{5}}{\dfrac{12}{5}\:-\:\dfrac{6}{5}}}$

$\displaystyle{\implies\sf\:\dfrac{\dfrac{12\:+\:6}{5}}{\dfrac{12\:-\:6}{5}}}$

$\displaystyle{\implies\sf\:\dfrac{\dfrac{18}{5}}{\dfrac{6}{5}}}$

$\displaystyle{\implies\sf\:\dfrac{18}{\cancel{5}}\:\times\:\dfrac{\cancel{5}}{6}}$

$\displaystyle{\implies\sf\:\cancel{\dfrac{18}{6}}}$

$\displaystyle{\implies\sf\:3}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:\dfrac{3\:\cos\:A\:+\:2\:\sin\:A}{3\:\cos\:A\:-\:2\:\sin\:A}\:=\:3\:}}}}$

Answer 2

Answer:

Step-by-step explanation:

Please the attached file

The answer is 3