Question

plz answer it......​

Answer 1

34.

3 (sin36/cos54)² - 2 (tan18 /cot72)² + 2tan13.tan21.tan69.tan77

Formulas used are ,

sin A= cos (90-A)

cos A= sin(90-A)

cot A=tan (90-A)

tan A = cot(90-A)

cot A × tan A = 1

so , sin 36 = cos (90-36) = cos 54

and, tan 18 = cot (90-18) = cot 72

and, tan 77 = cot (90-77) = cot 13

and, tan 69 = cot(90-69) = cot21

so , sin36/cos54 = 1

tan18/cot72 = 1

tan13× tan21× tan69× tan77 = tan13× cot13× tan21× cot21 = 1

=> 3(1)² - 2(1)² + 2×1 = 3-2+2 = 3

Solution of next question ,

P + Q + R = 180° (as P,Q,R are interior angles of a triangle)

=> (P+Q)/2 = (180°-R)/2 = 180°/2 - R/2 = 90°-R/2

taking sin both sides , we get

=> sin (P+Q/2) = sin (90° - R/2) = cos R/2

and

taking tan both sides , we get

=> tan (P+Q/2) = tan (90°-R/2) = cot R/2

proved.