RowQ
The Vault
RowQ
The Vault
CBSE Class 10 Maths · 11 questions · 26 marks
Trigonometry begins with a simple observation: in a right triangle, fixing one acute angle fixes the ratios of the sides, no matter how large you draw the triangle. Those six ratios — sine, cosine, tangent and their reciprocals — are the entire vocabulary of the chapter. Learn the standard-angle table until it is automatic, and treat identity proofs as tidying exercises where you convert everything to sin and cos and simplify.
If sin θ = 3/5 for an acute angle θ, then tan θ equals:
Answer
tan θ = 3/4. sin θ = opposite/hypotenuse = 3/5, so take the opposite side as 3 and the hypotenuse as 5. By Pythagoras, adjacent = √(5² - 3²) = √(25 - 9) = √16 = 4. Therefore tan θ = opposite/adjacent = 3/4.
The value of 2 tan²45° + cos²30° - sin²60° is:
Answer
The value is 2. tan 45° = 1, so 2 tan²45° = 2(1)² = 2. cos 30° = √3/2, so cos²30° = 3/4. sin 60° = √3/2, so sin²60° = 3/4. Adding: 2 + 3/4 - 3/4 = 2.
The expression (1 - cos²θ) cosec²θ simplifies to:
Answer
It simplifies to 1. From sin²θ + cos²θ = 1 we get 1 - cos²θ = sin²θ. Also cosec²θ = 1/sin²θ. So (1 - cos²θ) cosec²θ = sin²θ × (1/sin²θ) = 1, provided sin θ ≠ 0.
If 3 cot θ = 4 for an acute angle θ, then sec θ equals:
Answer
sec θ = 5/4. From 3 cot θ = 4 we get cot θ = 4/3, so tan θ = 3/4. Taking the opposite side as 3 and the adjacent side as 4, the hypotenuse is √(9 + 16) = 5. Then cos θ = adjacent/hypotenuse = 4/5, so sec θ = 1/cos θ = 5/4.
Assertion (A): There exists an acute angle θ for which sin θ = 5/4. Reason (R): For any acute angle θ, the value of sin θ lies between 0 and 1.
Answer
A is false but R is true. Reason R is correct: sin θ is the ratio of the side opposite θ to the hypotenuse, and since the hypotenuse is the longest side of a right triangle, this ratio can never exceed 1. The Assertion claims sin θ = 5/4, which is greater than 1 and therefore impossible. So A is false, and in fact R is the very reason A fails.
If tan θ = 5/12 for an acute angle θ, find sin θ and cos θ.
Answer
tan θ = opposite/adjacent = 5/12, so take the opposite side as 5 and the adjacent side as 12. By Pythagoras, hypotenuse = √(5² + 12²) = √(25 + 144) = √169 = 13. Therefore sin θ = 5/13 and cos θ = 12/13. Check: sin²θ + cos²θ = 25/169 + 144/169 = 169/169 = 1 ✔
Prove that (1 + tan²A)/(1 + cot²A) = tan²A.
Answer
Use the identities 1 + tan²A = sec²A and 1 + cot²A = cosec²A. LHS = sec²A/cosec²A = (1/cos²A) ÷ (1/sin²A) = (1/cos²A) × (sin²A/1) = sin²A/cos²A = tan²A = RHS. Hence proved, for all A where cos A ≠ 0 and sin A ≠ 0.
Evaluate sin 60° cos 30° + cos 60° sin 30°.
Answer
Substitute the standard values: sin 60° = √3/2, cos 30° = √3/2, cos 60° = 1/2, sin 30° = 1/2. sin 60° cos 30° = (√3/2)(√3/2) = 3/4. cos 60° sin 30° = (1/2)(1/2) = 1/4. Adding, 3/4 + 1/4 = 1. The value of the expression is 1.
Prove that (cos A)/(1 - tan A) + (sin A)/(1 - cot A) = sin A + cos A.
Answer
Step 1 — Rewrite each denominator in terms of sin A and cos A. 1 - tan A = 1 - sin A/cos A = (cos A - sin A)/cos A. 1 - cot A = 1 - cos A/sin A = (sin A - cos A)/sin A. Step 2 — Simplify each fraction. (cos A)/(1 - tan A) = cos A × cos A/(cos A - sin A) = cos²A/(cos A - sin A). (sin A)/(1 - cot A) = sin A × sin A/(sin A - cos A) = sin²A/(sin A - cos A) = -sin²A/(cos A - sin A). Step 3 — Add the two results over the common denominator (cos A - sin A): LHS = (cos²A - sin²A)/(cos A - sin A). Step 4 — Factorise the numerator as a difference of squares: cos²A - sin²A = (cos A - sin A)(cos A + sin A). Step 5 — Cancel the common factor (cos A - sin A), which is non-zero provided A ≠ 45°: LHS = cos A + sin A = RHS. Hence proved.
(a) Prove that √[(1 + sin θ)/(1 - sin θ)] = sec θ + tan θ. (b) If 5 tan θ = 4, find the value of (5 sin θ - 3 cos θ)/(5 sin θ + 2 cos θ).
Answer
(a) Multiply numerator and denominator inside the root by (1 + sin θ): (1 + sin θ)/(1 - sin θ) = (1 + sin θ)²/[(1 - sin θ)(1 + sin θ)] = (1 + sin θ)²/(1 - sin²θ). Since 1 - sin²θ = cos²θ, this equals (1 + sin θ)²/cos²θ. Taking the positive square root for an acute θ: √[(1 + sin θ)/(1 - sin θ)] = (1 + sin θ)/cos θ = 1/cos θ + sin θ/cos θ = sec θ + tan θ. Hence proved. (b) From 5 tan θ = 4 we get tan θ = 4/5. Divide the numerator and the denominator of the given expression by cos θ: (5 sin θ - 3 cos θ)/(5 sin θ + 2 cos θ) = (5 tan θ - 3)/(5 tan θ + 2). Substituting 5 tan θ = 4: = (4 - 3)/(4 + 2) = 1/6. The value is 1/6.
Read the following and answer the questions that follow: A carpenter builds a triangular support bracket for a shelf. The bracket is a right triangle ABC, right-angled at B, in which AB = 24 cm lies flat along the wall's horizontal line and BC = 7 cm rises vertically. The acute angle at vertex A is called θ. (a) Find the length of the sloping edge AC. (b) Write the values of sin θ and cos θ. (c) Verify that sin²θ + cos²θ = 1 for this bracket. (d) Find tan θ and verify that 1 + tan²θ = sec²θ.
Answer
(a) By Pythagoras Theorem, AC² = AB² + BC² = 24² + 7² = 576 + 49 = 625, so AC = 25 cm. (b) With respect to θ at vertex A, the opposite side is BC = 7 and the hypotenuse is AC = 25, while the adjacent side is AB = 24. sin θ = 7/25 and cos θ = 24/25. (c) sin²θ + cos²θ = (7/25)² + (24/25)² = 49/625 + 576/625 = 625/625 = 1 ✔ (d) tan θ = opposite/adjacent = 7/24, so tan²θ = 49/576 and 1 + tan²θ = (576 + 49)/576 = 625/576. Also sec θ = 1/cos θ = 25/24, so sec²θ = 625/576. The two agree, verifying 1 + tan²θ = sec²θ.
RowQ generates fresh questions on Introduction to Trigonometry, marks your answers, and explains every step.
Start free