Question

If Cos2a-sin2a=tan2b,prove that √2cosa.cosb=1

Answer 1

We start with:

cos² a - sin² a = tan² b

and from the trigonometric formula: cos² a + sin² a = 1

becomes  : sin² a = 1 - cos² a

Also we know that   tan b = sin b / cos b

Therefore: cos² a - ( 1 - cos² a ) = sin² b / cos² b  

cos² a - 1 + cos² a = sin² b / cos² b

2 cos² a - 1 = ( 1 - cos² b ) / cos² b     /· ( cos²b )  - we have to multiply both sides of the equation by cos²b

2 cos² a cos² b - cos² b = 1 - cos² b

2 cos² a cos² b = 1 - cos² b + cos² b

2 cos² a cos² b = 1   ( take the square root of the both sides of the equation )

√2 cos a · cos b = 1

Hence proved.

Answer 2

given, cos²a - sin²a = tan²b
we have to prove that √2cosa.cosb = 1

cos²a - sin²a = tan²b

cos²a - (1 - cos²a) = tan²b

cos²a - 1 + cos²a = tan²b

2cos²a = 1 + tan²b

we know, sec²x - tan²x = 1,
so, 1 + tan²b = sec²b

2cos²a = sec²b

2cos²a = 1/cos²b

2cos²a.cos²b = 1

taking square root both sides ,

√2cosa.cosb = ±1

hence proved √2cosa.cosb = 1