Q) Prove that Ans: Let’s start from LHS LHS = = = We know that sin2A + cos2A = 1 LHS = = =tan A …….. RHS Hence proved!
Prove that: (sin A – 2 sin^3 A)/(2cos^3 A – cos A) = tan A Read More »
trigonometry
Q) If A and B are acute angles such that sin(A-B) = 0 and 2 cos (A+B) -1 = 0, then find angles A and B. Ans:Â Given that sin(A – B) = 0 Since we know that, sin 00 = 0Â Â Â Â Â Â Â A – B = 0 ……….. (i)
Q) Evaluate Ans: Since sin2θ + cos2θ = 1        = = 5 cos2 60 + 4 sec2 30 – tan2 45 = = = =
Q) Prove that: = 1 + sinθ cosθ Ans: Let’s start from LHS LHS = = = = = We know that, a3 – b3 = (a – b) (a2 + b2 + ab) LHS = = (cos2 θ + sin2 θ + sin θ cos θ) = 1 + sinθ cosθ = RHS …………. Hence
Prove that: cos^2 θ/(1 – tan θ) + sin^3 θ/(sin θ – cos θ) = 1 + sin θ cos θ Read More »
Q) Prove that: = 2 cosec θ Ans: LHS = = = = = = = = 2 cosecθ = RHS ……… Hence Proved !
Prove that: sin θ / (1 + cos θ) + (1 + cos θ) / sin θ = 2 cosec θ Read More »
Q) Prove that: x Ans: Let’s start from LHS LHS = x = x =  x = sin θ x cos θ Now, take RHS RHS = = = = sin θ x cos θ Since LHS = RHS Hence proved!
Prove that: (1/cosθ – cos θ) (1/(sin θ) – sin θ) = 1/(tan θ+ cot θ) Read More »
Q) If cos A + cos2 A = 1, then find the value of sin2 A + sin4 A. Ans: cos A + cos2 A = 1 cos A = 1 – cos2 A cos A = sin2 A…………. (i) Now, sin2 A + sin4 A = (sin2 A) + (sin2 A) 2 by substituting
If Cos A + Cos^2 A = 1, then find the value of Sin^2 A + Sin^4 A. Read More »
Q) If 4 cot2 45° – sec2 60° + sin2 60° + p =  then find the value of p. Ans: VIDEO SOLUTION STEP BY STEP SOLUTION 4 cot2 45° – sec2 60° + sin2 60° + p = 4 (1)2 – (2)2 + 4 – 4 + + p = p = 0 Therefore,
If 4 cot^2 45° — sec2^ 60° + sin^2 60° + p = 3/4 then find the value of p. Read More »
Q) Prove that, 2(sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) +1 = 0 Ans: We know that, (a-b)3 = a3 – b3 – 3ab (a – b) And (a-b)2 = a2 + b2 – 2ab LHS: 2(sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) +1 = 2[sin6
Prove that, 2(sin^6 θ + cos^6 θ) – 3 (sin^4 θ + cos^4 θ) +1 = 0 Read More »
Q) If tan θ = , then show that  = Ans: Given that, tan θ =   cot θ = √7 Let’s start from numerator of LHS: cosec2 θ – sec2 θ = (1 + cot2 θ) – (1 + tan2 θ) = cot2 θ – tan2 θ = (√7)2 – = 7 – = …………………