Question

Question 9 cos(3π/2 + x) . cos(2π + x) [ cot(3π/2 - x) . cot(2π + x) ] = 1

Class X1 - Maths -Trigonometric Functions Page 73

Answer 1

LHS = cos(3π/2 + x).cos(2π+x)[cot(3π/2 -x) + cot(2π+ x) ]

we know,
cos(3π/2 + x) = sinx
cos(2π+x) = cosx
cot(3π/2 -x) = tanx
cot(2π+x) = cotx, use this here,

= sinx.cosx[tanx + cotx ]
= sinx.cosx [sinx/cosx + cosx/sinx]
= sinx.cosx[(sin²x + cos²x]/sinx.cosx
= (sin²x + cos²x ) = 1 = RHS

Answer 2

Answer:

Hi, By using identity sin2x + cos2x =1.

Step-by-step explanation:

Your answer is in the above photograph.