Question

If 3(cos^2 25° +cos^2 65°)/ (sin^2 27°+sin^2 63°)=a/3 then find value of a

Answer 1

Please see the attachment

Answer 2

Answer:

Numeric value of a is 9.

Step-by-step explanation:

Given,

$\dfrac{3(cos^2 25\degree + cos^2 65\degree)}{sin^2 27\degree + sin^2 63\degree} = \dfrac{a}{3}$

From the trigonometric identities, we know,

cos^2 A = cos^2{ 90 - ( 90 - A ) }

cos^2 ( 90 - A ) = sin^2 A

sin^2 A = sin^2{ 90 - ( 90 - A ) }

sin^2 ( 90 - A ) = cos^2

sin^2 A + cos^2 A = 1

So,

$\implies \dfrac{3(cos^2 25\degree + cos^2[ 90 - (90 - 65) ] )}{sin^2 27\degree + sin^2[ 90 - (90-63)]}=\dfrac{a}{3}$

$\implies \dfrac{3(cos^2 25\degree + cos^2(90 - 25))}{sin^2 27\degree + sin^2(90-27)}=\dfrac{a}{3}$

$\implies \dfrac{3(cos^2 25\degree + sin^2 25)}{sin^2 27\degree + cos^2 27\degree } = \dfrac{a}{3}$

$\implies \dfrac{3(1)}{1} =\dfrac{a}{3}$

$\implies 3 \times 3 = a$

$\implies 9 = a$

Therefore the value of a is 9.