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

Tip: As x belongs in the 1st quadrant, then all the trigonometric ratios are positive.

Answer: (sin x + cos x)/(sin x - cos x) = - 7.

Explanation:

tan x = 3/4

Here, (sin x + cos x)/(sin x - cos x)

= [(sin x + cos x)/sin x]/[(sin x - cos x)/sin]

(Dividing numerator & denominator by sin x.)

= (1 + cot x)/(1 - cot x)

= (1 + 4/3)/(1 - 4/3)

(As, tan^-1 x = cot x.)

= (7/3)/(- 1/3)

= - 7.

Answer 2

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

Step-by-step-explanation

We have given that,

$\displaystyle{\sf\:\tan\:x\:=\:\dfrac{3}{4}} \: \sf \: {,x \: lies \: in \: the \: first \: quadrant .}$

We have to find the value of

$\displaystyle{\sf\:\dfrac{\sin\:x\:+\:\cos\:x}{\sin\:x\:-\:\cos\:x}} </p><p>$

Now,

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

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

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

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

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

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

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

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