Question

"Question33
Chapter6,T-Ratios of particular angles Exercise -6 ,Page number 289
If A = B = 45°,verify that : cos (A + B) = cos A cos B - sin A sin B "

Answer 1

Hey there! Thanka for the question!

Given,
A = B = 45°

Now,
cos(A + B) = cosA cosB - sinA sinB

Substituting A = 45 , B = 45 °

cos( 45 + 45 ) = cos45 cos45 - sin45 sin45

cos90 = cos²45 - sin²45

We know, Trigonometry ratios of particular angles : cos90 = 0 , cos45 = 1/√2 , sin45 = 1/√2

0= (1/√2)² - (1/√2)²

0 = 1/2 - 1/2

0 = 0

Both Sides of the equation are equal.

Hence, We proved and verified that cos (A + B) = cos A cos B - sin A sin B holds good for A = B = 45°

Answer 2

Given that

A = B = 45°

To verify

cos ( A + B ) = cos A cos B - sin A sin B

substitute the value of A and B

cos ( 45 + 45 ) = cos 45 cos 45 - sin 45 cos 45

On RSH

$\frac{1}{ \sqrt{2} } . \frac{1}{ \sqrt{2} } - \frac{1}{ \sqrt{2} } . \frac{1}{ \sqrt{2} }$

= 1/2 - 1 /2

= 0

ON LHS

cos 90 = 0

So LHS = RHS

Hence proved