Question

cot(45°-A)=tan 2A+sec 2A prove it​

Answer 1

Answer:

L.H.S.

= 1 / cos(2a) + sin(2a) / cos(2a)

= (1 + sin(2a)) / (cos(2a))

= (sin^2(a) + cos^2(a) + 2 * sin(a) * cos(a)) / (cos^2(a) - sin^2(a))

= (sin(a) + cos(a))^2 / ((cos(a) - sin(a)) * (cos(a) + sin(a)))

= (cos(a) + sin(a)) / (cos(a) - sin(a))

= cos(a) * (1 + sin(a) / cos(a)) / (cos(a) * (1 - sin(a)/cos(a)))

= (1 + tan(a)) / (1 - tan(a))

= (tan(45) + tan(a)) / (1 - tan(a) * tan(45))

= tan(a + 45)

Answer 2

Step-by-step explanation:

cot(45 -A) = tan2·(A) + sec2·(A)

We have;

tan2(A) + sec2(A) = sin2(A)/cos2(A) + 1/cos2(A) = (sin2(A) + 1)/cos2(A)

= (sin²(A) + cos²(A) + 2·sin·(A)cos(A))/(cos²A - sin²A)

= (sin(A) + cos(A))²/((cos(A) -sin(A))(cos(A) +sin(A)))

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

= (cos(A)(1 + sin(A)/cos(A)))/(cos(A)(1 - sin(A)/cos(A)))

= (1 + tan(A))/(1 - tan(A))

= (tan 45 + tan(A))/(1 - tan(45)·tan(A)) = tan(45 + A)

cot(π/2 - x) = tan(x)

= cot(π/2 - (45 + A)) = tan(45 + A)

π/2 = 90°

= cot(90 - (45 + A)) = tan(45 + A)

cot(45 - A) = tan(45 + A) = tan(2A) + sec(2A).