Question

if tanA = b/a then show that ( a cos 2A + b sin 2A) = a​

Answer 1

Question : -

If tan A = b/a the show that (a cos 2A + b sin 2A) = a

ANSWER

Given : -

tan A = b/a

Required to prove : -

• a cos 2A + b sin 2A = a

Proof : -

tan A = b/a

So, let's consider a right angled ∆ABC, in which angle B = 90°

In ∆ABC, right angled at B

tan A = b/a (given)

But,

tan A = (opposite side)/(adjacent side)

tan A = (BC)/(AB)

So,

BC = bk units & AB = ak units

Applying Pythagoras theorem;

AC² = AB² + BC²

AC² = (ak)² + (bk)²

AC² = a²k² + b²k²

AC² = k²(a²+b²)

AC = √[k²(a²+b²]

AC = √(a²+b²)k units

Now, let's evaluate the value of sin A & cos A

sin A = (opposite side)/(hypotenuse)

sin A = (BC)/(AC)

sin A = (bk)/(√[a²+b²]k)

sin A = (b)/(√[a²+b²])

Similarly,

cos A = (adjacent side)/(hypotenuse)

cos A = (AB)/(AC)

cos A = (ak)/(√[a²+b²]k)

cos A = (a)/(√[a²+b²])

Now, we know that

• sin 2A = 2 sin x cos x
• sin 2A = 2 sin x cos xcos 2A = cos² A - sin² A

Let's find the value of (a cos 2A + b sin 2A)

a cos 2A + b sin 2A

a (cos² A - sin² A) + b (2 sin A cos A)

a [(a)/(√a²+b²)]²-[(b)/(√a²+b²)]² + b ( [2] x [(b)/(√a²+b²)]² x [(a)/(√a²+b²)]² )

a [ (a² - b²)/(a²+b²) ] + b [ (2ab)/(a²+b²) ]

(a³-ab²)/(a²+b²) + (2ab²)/(a²+b²)

(a³-ab²+2ab²)/(a²+b²)

(a³+ab²)/(a²+b²)

(a[a²+b²])/(a²+b²)

= a

= RHS

Hence Proved !…

Answer 2

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