Exercise 8.3 — Introduction to Trigonometry

1. 13 marksNCERT Cl-10 Maths, Ex 8.3, Q1

   Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

   Hint. The identity that contains cot is cosec²A = 1 + cot²A. Get cosec A from it first, and everything else follows.

   Step 1 — Start from the identity that involves cot. cosec²A = 1 + cot²A, so cosec A = √(1 + cot²A)

   We take the positive root because A is acute here, and all six ratios are positive for acute angles.

   Step 2 — Get sin A, which is the reciprocal of cosec A.

   sin A = 1 / √(1 + cot²A)

   Step 3 — Get tan A, the reciprocal of cot A.

   tan A = 1 / cot A

   Step 4 — Get sec A. Use sec²A = 1 + tan²A with tan A = 1/cot A: sec²A = 1 + 1/cot²A = (cot²A + 1)/cot²A

   sec A = √(1 + cot²A) / cot A

   Check with A = 45°, where cot A = 1: sin A = 1/√2 ✓, tan A = 1 ✓, sec A = √2 ✓

   ✦ Answer: sin A = 1/√(1 + cot²A), tan A = 1/cot A, sec A = √(1 + cot²A)/cot A

   Where students slip. Writing sin A = √(1 − cos²A) and calling it an answer in terms of cot. The question fixes which ratio you may use, so every symbol on the right must be cot A.

   Another way. Build a triangle instead: cot A = adjacent/opposite means adjacent = cot A and opposite = 1, so the hypotenuse is √(1 + cot²A). Then read all three ratios straight off.

2. 24 marksNCERT Cl-10 Maths, Ex 8.3, Q2

   Write all the other trigonometric ratios of ∠A in terms of sec A.

   Hint. sec²A − tan²A = 1 gives you tan A immediately. cos A is a reciprocal, and the rest follow.

   Step 1 — cos A is the reciprocal of sec A.

   cos A = 1 / sec A

   Step 2 — Get tan A from the identity sec²A = 1 + tan²A. tan²A = sec²A − 1, so

   tan A = √(sec²A − 1)

   Step 3 — Get sin A. Since tan A = sin A / cos A, we have sin A = tan A × cos A: sin A = √(sec²A − 1) × (1/sec A)

   sin A = √(sec²A − 1) / sec A

   Step 4 — The remaining two are reciprocals of what we already have.

   cosec A = sec A / √(sec²A − 1) cot A = 1 / √(sec²A − 1)

   Check with A = 60°, where sec A = 2: cos A = 1/2 ✓, tan A = √3 ✓, sin A = √3/2 ✓, cosec A = 2/√3 ✓, cot A = 1/√3 ✓

   ✦ Answer: cos A = 1/sec A; sin A = √(sec²A − 1)/sec A; tan A = √(sec²A − 1); cosec A = sec A/√(sec²A − 1); cot A = 1/√(sec²A − 1)

   Where students slip. Deriving sin A as √(1 − cos²A) and leaving cos A in the answer. Substitute the whole way down until only sec A remains.

   Another way. The triangle route is quicker: sec A = hypotenuse/adjacent = sec A / 1, so adjacent = 1 and hypotenuse = sec A, making the opposite side √(sec²A − 1). Every ratio is then a reading.

3. 34 marksNCERT Cl-10 Maths, Ex 8.3, Q3

   Choose the correct option and justify. (i) 9 sec²A − 9 tan²A = (A) 1 (B) 9 (C) 8 (D) 0. (ii) (1 + tan θ + sec θ)(1 + cot θ − cosec θ) = (A) 0 (B) 1 (C) 2 (D) −1. (iii) (sec A + tan A)(1 − sin A) = (A) sec A (B) sin A (C) cosec A (D) cos A. (iv) (1 + tan²A)/(1 + cot²A) = (A) sec²A (B) −1 (C) cot²A (D) tan²A.

   Hint. Each one collapses to a single identity. Take out common factors first, and convert to sin and cos when nothing else suggests itself.

   (i) Take out the 9. 9 sec²A − 9 tan²A = 9(sec²A − tan²A)

   The bracket is exactly the second identity, which equals 1.

   = 9 × 1 = 9

   ✦ (B) 9

   (ii) Convert everything to sin and cos over a common denominator. 1 + tan θ + sec θ = (cos θ + sin θ + 1)/cos θ 1 + cot θ − cosec θ = (sin θ + cos θ − 1)/sin θ

   Multiplying: = [(cos θ + sin θ) + 1][(cos θ + sin θ) − 1] / (sin θ cos θ)

   The numerator is a difference of squares: = [(cos θ + sin θ)² − 1] / (sin θ cos θ) = [cos²θ + 2 sin θ cos θ + sin²θ − 1] / (sin θ cos θ)

   Since cos²θ + sin²θ = 1, the 1 and the −1 cancel: = 2 sin θ cos θ / (sin θ cos θ) = 2

   ✦ (C) 2

   (iii) Write the bracket over cos A. sec A + tan A = (1 + sin A)/cos A

   So the product is (1 + sin A)(1 − sin A)/cos A = (1 − sin²A)/cos A = cos²A/cos A = cos A

   ✦ (D) cos A

   (iv) Replace each bracket using an identity. 1 + tan²A = sec²A and 1 + cot²A = cosec²A, so the fraction is sec²A / cosec²A = (1/cos²A) × (sin²A/1) = sin²A/cos²A = tan²A

   ✦ (D) tan²A

   Where students slip. In (i), reading 9 sec²A − 9 tan²A as 9(sec²A − tan²A) but then writing sec² − tan² = 0 by analogy with sin² + cos² = 1. The identity is sec²A − tan²A = **1**, so the answer is 9, not 0.

   Another way. Every one of these can be spot-checked with a convenient angle. At θ = 45°, part (ii) reads (1 + 1 + √2)(1 + 1 − √2) = (2 + √2)(2 − √2) = 4 − 2 = 2 ✓, confirming the option in one line.

4. 410 marksNCERT Cl-10 Maths, Ex 8.3, Q4

   Prove the following identities, the angles being acute and the expressions defined. (i) (cosec θ − cot θ)² = (1 − cos θ)/(1 + cos θ); (ii) cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A; (iii) tan θ/(1 − cot θ) + cot θ/(1 − tan θ) = 1 + sec θ cosec θ; (iv) (1 + sec A)/sec A = sin²A/(1 − cos A); (v) (cos A − sin A + 1)/(cos A + sin A − 1) = cosec A + cot A, using cosec²A = 1 + cot²A; (vi) √((1 + sin A)/(1 − sin A)) = sec A + tan A; (vii) (sin θ − 2 sin³θ)/(2 cos³θ − cos θ) = tan θ; (viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan²A + cot²A; (ix) (cosec A − sin A)(sec A − cos A) = 1/(tan A + cot A); (x) (1 + tan²A)/(1 + cot²A) = ((1 − tan A)/(1 − cot A))² = tan²A.

   Hint. Take one side and transform it. Converting to sin and cos is the universal fallback, but several of these are quicker with a difference of squares or a common factor.

   (i) LHS = (cosec θ − cot θ)² = ((1 − cos θ)/sin θ)² = (1 − cos θ)²/sin²θ Replace sin²θ by 1 − cos²θ = (1 − cos θ)(1 + cos θ): = (1 − cos θ)² / [(1 − cos θ)(1 + cos θ)] = (1 − cos θ)/(1 + cos θ) = RHS ✓

   (ii) Put the two fractions over the common denominator cos A(1 + sin A): LHS = [cos²A + (1 + sin A)²] / [cos A(1 + sin A)] = [cos²A + 1 + 2 sin A + sin²A] / [cos A(1 + sin A)] Since cos²A + sin²A = 1, the top is 2 + 2 sin A = 2(1 + sin A): = 2(1 + sin A)/[cos A(1 + sin A)] = 2/cos A = 2 sec A = RHS ✓

   (iii) Write everything in sin and cos. With cot θ = cos θ/sin θ, first term = (sin θ/cos θ)/((sin θ − cos θ)/sin θ) = sin²θ/[cos θ(sin θ − cos θ)] second term = (cos θ/sin θ)/((cos θ − sin θ)/cos θ) = cos²θ/[sin θ(cos θ − sin θ)] The two denominators differ by a sign, so combining over sin θ cos θ (sin θ − cos θ): = (sin³θ − cos³θ)/[sin θ cos θ(sin θ − cos θ)] Factor the difference of cubes: sin³θ − cos³θ = (sin θ − cos θ)(sin²θ + sin θ cos θ + cos²θ) = (sin²θ + sin θ cos θ + cos²θ)/(sin θ cos θ) = (1 + sin θ cos θ)/(sin θ cos θ) = 1 + 1/(sin θ cos θ) = 1 + sec θ cosec θ = RHS ✓

   (iv) LHS = (1 + sec A)/sec A = cos A(1 + sec A) = cos A + 1 RHS = sin²A/(1 − cos A) = (1 − cos²A)/(1 − cos A) = (1 − cos A)(1 + cos A)/(1 − cos A) = 1 + cos A The two sides agree ✓ — this is the one where working each side separately is much easier than transforming one into the other.

   (v) Divide the top and bottom of the LHS by sin A, so that cosec and cot appear: LHS = (cot A − 1 + cosec A)/(cot A + 1 − cosec A) Use 1 = cosec²A − cot²A = (cosec A − cot A)(cosec A + cot A) in the numerator: numerator = (cot A + cosec A) − (cosec A − cot A)(cosec A + cot A) = (cot A + cosec A)[1 − (cosec A − cot A)] = (cot A + cosec A)(1 − cosec A + cot A) The second bracket is exactly the denominator, so it cancels: = cosec A + cot A = RHS ✓

   (vi) Multiply inside the root by (1 + sin A)/(1 + sin A): LHS = √[(1 + sin A)²/((1 − sin A)(1 + sin A))] = √[(1 + sin A)²/(1 − sin²A)] = √[(1 + sin A)²/cos²A] = (1 + sin A)/cos A = 1/cos A + sin A/cos A = sec A + tan A = RHS ✓

   (vii) Take out the common factors on top and bottom: LHS = sin θ(1 − 2 sin²θ) / [cos θ(2 cos²θ − 1)] Now 1 − 2 sin²θ = 1 − 2(1 − cos²θ) = 2 cos²θ − 1, so the two brackets are the same and cancel: = sin θ/cos θ = tan θ = RHS ✓

   (viii) Expand both squares: (sin A + cosec A)² = sin²A + 2 sin A cosec A + cosec²A = sin²A + 2 + cosec²A (cos A + sec A)² = cos²A + 2 cos A sec A + sec²A = cos²A + 2 + sec²A (each middle term is 2, because a ratio times its reciprocal is 1) Adding: (sin²A + cos²A) + 4 + cosec²A + sec²A = 1 + 4 + (1 + cot²A) + (1 + tan²A) = 7 + tan²A + cot²A = RHS ✓

   (ix) LHS = (1/sin A − sin A)(1/cos A − cos A) = [(1 − sin²A)/sin A][(1 − cos²A)/cos A] = (cos²A/sin A)(sin²A/cos A) = sin A cos A RHS = 1/(tan A + cot A) = 1 / [(sin A/cos A) + (cos A/sin A)] = 1 / [(sin²A + cos²A)/(sin A cos A)] = sin A cos A Both sides equal sin A cos A ✓

   (x) Left expression: (1 + tan²A)/(1 + cot²A) = sec²A/cosec²A = sin²A/cos²A = tan²A Middle expression: rewrite cot A as 1/tan A, so (1 − tan A)/(1 − 1/tan A) = (1 − tan A) × tan A/(tan A − 1) = −tan A Squaring gives (−tan A)² = tan²A So all three expressions equal tan²A ✓

   ✦ Answer: all ten identities are established.

   Where students slip. Cross-multiplying an identity and working on both sides at once. That assumes the result is already true. Fix one side, transform it step by step, and arrive at the other.

   Another way. Two habits cover almost all of these: convert to sin and cos when stuck, and look for a difference of squares to multiply by — parts (i) and (vi) both fall to multiplying by the conjugate. Numerically spot-checking each identity at, say, 37° is a fast way to catch an algebra slip before writing the proof out.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 10 Mathematics textbook, Reprint 2026-27 (chapter 8 now runs to three exercises — 8.1, 8.2 and 8.3; trigonometric ratios of complementary angles were removed from this chapter, and the summary's six points make no mention of them). Questions are referenced from the NCERT textbook for identification.