Question

solve pls and help me... ​

Answer 1

Solution

Let, theta = x

Given:-

• tan x = p/q

Find:-

• (cos x + sin x )/(cos x - sin x)

To prove:-

• (p sin x - q cos x)/(p sin x + q cos x) = (p²-q²)/(p²+q²

Explanation

We know,

☛ tan x = (perpendicular )/(Base) = p/q

By, Pythagoras Theorem

☞ (Hypotenuse )² = (perpendicular)²+ (Base)²

➠ (Hypotenuse) = √[(p²+q²)]

Now, calculate other trigonometry ratio

☞ sin x = (perpendicular)/(Hypotenuse)

☞ sin x = p/√(p²+q²)

☞ cos x = (Base)/(Hypotenuse)

☞ cos x = q/√(p²+q²)

First calculate,

☞ (cos x + sin x )/(cos x - sin x)

keep above value,

➠[ {q/√(p²+q²)} + { p/√(p²+q²)}]/[{q/√(p²+q²} - p/√(p²+q²)}]

➠ [(p+q)/√(p²+q²)] / [ (q-p)/√(p²+q²)]

➠ [(p+q)/√(p²+q²)] × [√(p²+q²)/(q-p)]

➠ (p+q)/(q-p). [ Ans]

____________________

Prove

Take L.H.S.,

☞ (p sin x - q cos x)/(p sin x + q cos x)

keep value of cos x and sin x

➠ [ p . p/√(p²+q²) - q. q/√(p²+q²] / [ p. p/√(p²+q²) + q. q/√(p²+q²)]

➠ [(p²-q²)/√(p²+q²)] / [ (p²+q²)/√(p²+q²)]

➠ [(p²-q²)/√(p²+q²)] × [√(p²+q²)/(p²+q²)]

➠ [(p²-q²)/(p²+q²)]

= R.H.S

that's proved.