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Q) If tan θ =24/7, then find the value of sin θ + cos θ. (Q 22 A – 30/1/3 – CBSE 2026 Question Paper) Ans: Step 1: We know that in a right angled triangle, tan θ is given by: Since we are given that tan θ = ∴ We can assume that Opposite […]
If tan θ =24/7, then find the value of sin θ + cos θ. Read More »
Q) If cot θ =7/8, then find the value of (1+sin θ)(1−sin θ)/(1+cos θ)(1−cos θ). (Q 22 B – 30/1/3 – CBSE 2026 Question Paper) Ans: Step 1: Let’s first simiplify the given expression: = ………… (i) Step 2: From Trigonometric identity, we know that sin 2 θ + cos 2 θ = 1 ∴ 1
If cot θ =7/8, then find the value of (1+sin θ)(1−sin θ)/(1+cos θ)(1−cos θ). Read More »
Q) If x = h + a cos θ, y = k + b sin θ, then prove that : (Q 26 A – 30/1/3 – CBSE 2026 Question Paper) Ans: Step 1: We are given: x = h + a cos θ ∴ x – h = a cos θ …………… (i) Also, we
Q) Prove that : (Q 26 B – 30/1/3 – CBSE 2026 Question Paper) Ans: Step 1: Let’s start from LHS: LHS = = = = = = Step 2: Since 1 + tan 2 A = sec 2 A ∴ sec 2 A – 1 = tan 2 A Step 3: By substituting this
Prove that : tan A/(1+sec A) −tan A/(1−sec A) = 2 cosec A. Read More »
Q) Express cos A and tan A in terms of sin A. (Q 25 B – 30/2/3 – CBSE 2026 Question Paper) Ans: Step 1: We know that according to trigonometric identity: sin 2 θ + cos 2 θ = 1 or for θ = A, we can say, sin 2 A + cos 2 A = 1 ∴
Q) If tan A = 4 / 3, find sin A and cos A. (Q 25 A – 30/2/3 – CBSE 2026 Question Paper) Ans: Step 1: We know that in a right angled triangle, tan θ is given by: Since we are given that tan A = ∴ We can assume that Opposite side
Q) Prove that: √(1 – sin θ)/(1 + sin θ) = sec θ – tan θ. (Q 24 B – 30/2/1 – CBSE 2026 Question Paper) Ans: Let’s start from LHS: LHS = Step 1: [Note: Since in RHS, we need to get cos θ in denominator, hence, we need to nullify + sign] ∴
Prove that: √(1 – sin θ)/(1 + sin θ) = sec θ – tan θ. Read More »
Q) If tan θ + 1 / tan θ = 2, find the value of tan^2 θ + 1/ tan^2 θ. (Q 24 A – 30/2/1 – CBSE 2026 Question Paper) Ans: Step 1: Given that, By squaring on both sides we get: ∴ Step 2: ∵ (a + b) 2 = a 2 + b 2 + 2 a b ∴
If tan θ + 1/tan θ = 2, find the value of tan^2 θ + 1/tan^2 θ Read More »
Q) Prove that: 1 + = cosec α (Q 22 B – 30/2/2 – CBSE 2026 Question Paper) Ans: Let’s start from LHS: LHS = 1 + Step 1: From Trigonometric identities, we know that 1 + cot 2 α = cosec 2 α ∴ cot 2 α = cosec 2 α – 1 Step 2: Let’s substitute the value of cot 2
Prove that: 1 + (cot^2 α)/(1 + cosec α) = cosec α Read More »
Q) Evaluate: (Q 22 A – 30/2/2 – CBSE 2026 Question Paper) Ans:Â Let’s solve numerator and denominator one by one and then combine: Step 1: Solving for numerator: (5 cos 2Â 60Â 0 + 4 sec 2Â 30Â 0 – tan 2Â 45Â 0) = 5 Â Â Â Â (by putting values) = 5 = = Step 2: Solving for
Evaluate: (5cos^2 60 + 4 sec ^ 2 30 – tan ^2 45) / (sin^2 30 + cos^2 30) Read More »