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If x tan 45o cos 60o = sin 60o cot 600 , then x is equal to, Given x tan 45o cos 60o = sin 60o cot 600, If Tan A + 1/ Tan A = 2, then the value of tan2 A + 1/tan2 A is, If θ and 2θ - 45o are acute angles such that sin θ = cos (2θ - 45o), then tan θ is equal to, If 16cot x = 12, then (sinx - cosx)/(sinx + cosx) equals, Now (sinx - cosx)/(sinx + cosx) = {(sinx - cosx)/sinx}/{(sinx + cosx)/sinx}, [divide by sinx in numerator and denominator], => (sinx - cosx)/(sinx + cosx) = (sinx/sinx - cosx/sinx)/(sinx/sinx + cosx/sinx), => (sinx - cosx)/(sinx + cosx) = (1 - cotx)/(1 + cotx), => (sinx - cosx)/(sinx + cosx) = (1 - 3/4)/(1 + 3/4), => (sinx - cosx)/(sinx + cosx) = {(4 - 3)/4}/{(4 + 3)/4}, => (sinx - cosx)/(sinx + cosx) = (1/4)/(7/4), If 16 cot x = 12, then (sinx - cosx)/(sinx + cosx) equals, => (sinx - cosx)/(sinx + cosx) = (16-12)/(16+12) {Apply compundendo and dividendo}, If 5θ and 4θ are acute angles satisfying sin 5θ = cos 4θ, then 2 sin 3θ - √3 tan 3θ is equal to, Now 2*sin 3θ - √3*tan 3θ = 2*sin(3*10) - √3*tan(3*10), If ABC is a right angle triangle at C then cos(A + B) is, Given, ABC is a right angle triangle at C, {cos(90 - A) * cos A}/tan A + cos2 (90 - A) is equals to, {cos(90 - A) * cos A}/tan A + cos2 (90 - A), = {sin A * cos A}/tan A + sin2 A {cos(90 - A) = sin A}, = {sin A * cos A}/(sin A/cos A) + sin2 A {tan A = sinA/cos A}, = 1 {since cos2 A + sin2 A = 1}, If tan θ = a/b , then (a sin θ + b cos θ)/(a sin θ - b cos θ) is equal to, Now, (a sin θ + b cos θ)/(a sin θ - b cos θ), = (a sin θ/cos θ + b cos θ/cos θ)/(a sin θ/cos θ - b cos θ/cos θ) {divide cos θ in numerator and denominator}. 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