Question

If 2sin(3x-15)° = √3, then find the value of sin(square) (2x+10)° +tan(square) (x+5)°

Answer 1

Solve the first part

$2sin (3x-15)^{o} = \sqrt{3}$

$sin (3x-15)^{o} = \sqrt{3}/2$

$sin (3x-15)^{o} = sin 60^{o}$

$3x - 15 = 60$

$3x = 75$

$x = 25$

Now substitute the value of 'x' in second part

$sin^{2} (2x+10)^{o} + tan^2(x+5)^o$

$sin^2 60 + tan^2 30$

$( \sqrt{3}/2)^2 + (1/ \sqrt{3})^2$

$3/4 + 1/3$

$= 13/12$

Answer 2

Refer the attachment for ur answer .

Thnx