Question

find the value of sinx + cos x / sin x - cos x ,if tan x = 3÷4, x lies in the 1st quadrant​

Answer 1

Answer:

$\displaystyle{\boxed{\red{\sf\:\dfrac{\sin\:x\:+\:\cos\:x}{\sin\:x\:-\:\cos\:x}\:=\:-\:7}}}$

Step-by-step-explanation

We have given that,

$\displaystyle{\sf\:\tan\:x\:=\:\dfrac{3}{4}}$,

x lies in the first quadrant.

We have to find the value of

$\displaystyle{\sf\:\dfrac{\sin\:x\:+\:\cos\:x}{\sin\:x\:-\:\cos\:x}}$

Now,

$\displaystyle{\sf\:\dfrac{\sin\:x\:+\:\cos\:x}{\sin\:x\:-\:\cos\:x}}$

$\displaystyle{\implies\sf\:\dfrac{\dfrac{\sin\:x\:+\:\cos\:x}{\cos\:x}}{\dfrac{\sin\:x\:-\:\cos\:x}{\cos\:x}}\:\qquad\dots\:[\:Dividing\:by\:\cos\:x\:]}$

$\displaystyle{\implies\sf\:\dfrac{\dfrac{\sin\:x}{\cos\:x}\:+\:\cancel{\dfrac{\cos\:x}{\cos\:x}}}{\dfrac{\sin\:x}{\cos\:x}\:-\:\cancel{\dfrac{\cos\:x}{\cos\:x}}}}$

$\displaystyle{\implies\sf\:\dfrac{\tan\:x\:+\:1}{\tan\:x\:-\:1}\:\qquad\:\dots\:\left[\:\because\:\dfrac{\sin\:x}{\cos\:x}\:=\:\tan\:x\:\right]}$

$\displaystyle{\implies\sf\:\dfrac{\dfrac{3}{4}\:+\:1}{\dfrac{3}{4}\:-\:1}}$

$\displaystyle{\implies\sf\:\dfrac{\dfrac{3\:+\:4}{4}}{\dfrac{3\:-\:4}{4}}}$

$\displaystyle{\implies\sf\:\dfrac{\dfrac{7}{4}}{\dfrac{-\:1}{4}}}$

$\displaystyle{\implies\sf\:\dfrac{7}{\cancel{4}}\:\times\:\dfrac{\cancel{4}}{-\:1}}$

$\displaystyle{\implies\sf\:-\:7}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:\dfrac{\sin\:x\:+\:\cos\:x}{\sin\:x\:-\:\cos\:x}\:=\:-\:7}}}}$

Answer 2

Step-by-step explanation:

the answer is in the picture