Trigonometry class 10 exercise 8.3 solutions

1. tan A:
tan A = 1cot A

2. sec A:
Using the identity sec2A = 1 + tan2A:
sec²A = 1 + (1cot A)² = 1 + 1cot²A = cot²A + 1cot²A
sec A = √(cot2A + 1)cot A

3. sin A:
Using the identity cosec2A = 1 + cot2A:
cosec A = √(1 + cot²A)
Since sin A = 1 / cosec A,
sin A = 1√(1 + cot2A)

1. cos A: cos A = 1sec A

2. sin A: From sin2A + cos2A = 1,
sin²A = 1 – cos²A = 1 – 1sec²A = sec²A – 1sec²A
sin A = √(sec2A - 1)sec A

3. cosec A: cosec A = 1/sin A = sec A√(sec2A - 1)

4. tan A: From tan2A = sec2A - 1,
tan A = √(sec2A - 1)

5. cot A: cot A = 1/tan A = 1√(sec2A - 1)

(i) 9 sec²A − 9 tan²A = ?
= 9 (sec²A − tan²A)
Using the identity sec2A − tan2A = 1,
= 9 (1) = 9.
Answer: (B) 9

(ii) (1 + tanθ + secθ)(1 + cotθ − cosecθ) = ?
Convert to sin and cos: = (cosθ + sinθ + 1cosθ)(sinθ + cosθ – 1sinθ)
Numerator is ((sinθ + cosθ) + 1)((sinθ + cosθ) – 1) = (sinθ + cosθ)² – 1²
= sin²θ + cos²θ + 2sinθcosθ – 1 = 1 + 2sinθcosθ – 1 = 2sinθcosθ.
The expression is 2sinθcosθsinθcosθ = 2.
Answer: (C) 2

(iii) (sec A + tan A)(1 − sin A) = ?
= (1cos A + sin Acos A)(1 – sin A) = 1 + sin Acos A(1 – sin A)
= 1 – sin²Acos A = cos²Acos A = cos A.
Answer: (D) cos A

(iv) 1 + tan²A1 + cot²A = ?
= sec²Acosec²A = 1/cos²A1/sin²A = sin²Acos²A = tan²A.
Answer: (D) tan²A

(i) (cosecθ − cotθ)² = 1 − cosθ1 + cosθ
LHS = (1sinθ − cosθsinθ)² = (1 – cosθsinθ)²
= (1 – cosθ)²sin²θ = (1 – cosθ)²1 – cos²θ = (1 – cosθ)(1 – cosθ)(1 – cosθ)(1 + cosθ) = RHS. (Proved)

(ii) cos A1 + sin A + 1 + sin Acos A = 2 sec A
LHS = cos²A + (1 + sin A)²(1 + sin A)cos A
= cos²A + 1 + 2sinA + sin²A(1 + sin A)cos A = (cos²A + sin²A) + 1 + 2sinA(1 + sin A)cos A
= 1 + 1 + 2sinA(1 + sin A)cos A = 2(1 + sinA)(1 + sin A)cos A = 2cos A = 2 sec A = RHS. (Proved)

(iii) tanθ1 − cotθ + cotθ1 − tanθ = 1 + secθcosecθ
LHS = sinθ/cosθ1 – cosθ/sinθ + cosθ/sinθ1 – sinθ/cosθ
= sin²θcosθ(sinθ – cosθ) – cos²θsinθ(sinθ – cosθ)
= sin³θ – cos³θsinθcosθ(sinθ – cosθ)
= (sinθ – cosθ)(sin²θ + cos²θ + sinθcosθ)sinθcosθ(sinθ – cosθ)
= 1 + sinθcosθsinθcosθ = 1sinθcosθ + 1 = 1 + secθcosecθ = RHS. (Proved)

(iv) 1 + sec Asec A = sin²A1 − cos A
LHS = 1 + 1/cos A1/cos A = (cos A + 1)/cos A1/cos A = 1 + cos A.
RHS = 1 – cos²A1 – cos A = (1 – cos A)(1 + cos A)1 – cos A = 1 + cos A.
Since LHS = RHS, the identity is proved.

(v) cos A − sin A + 1cos A + sin A − 1 = cosec A + cot A
Divide numerator and denominator by sin A:
LHS = cot A – 1 + cosec Acot A + 1 – cosec A = (cot A + cosec A) – 1cot A – cosec A + 1
Replace 1 in numerator with (cosec²A – cot²A):
= (cot A + cosec A) – (cosec A – cot A)(cosec A + cot A)cot A – cosec A + 1
= (cosec A + cot A)[1 – (cosec A – cot A)]cot A – cosec A + 1
= (cosec A + cot A)[1 – cosec A + cot A]1 – cosec A + cot A = cosec A + cot A = RHS. (Proved)

(vi) √1 + sin A1 – sin A = sec A + tan A
LHS = √(1 + sin A)(1 + sin A)(1 – sin A)(1 + sin A) = √(1 + sin A)²1 – sin²A
= √(1 + sin A)²cos²A = 1 + sin Acos A = 1cos A + sin Acos A = sec A + tan A = RHS. (Proved)

(vii) sinθ − 2sin³θ2cos³θ − cosθ = tanθ
LHS = sinθ(1 – 2sin²θ)cosθ(2cos²θ – 1)
Replace 1 in numerator with sin²θ + cos²θ:
= tanθ sin²θ + cos²θ – 2sin²θ2cos²θ – 1 = tanθ cos²θ – sin²θ2cos²θ – (sin²θ + cos²θ)
= tanθ cos²θ – sin²θcos²θ – sin²θ = tanθ = RHS. (Proved)

(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan²A + cot²A
LHS = (sin²A + 2sinAcosecA + cosec²A) + (cos²A + 2cosAsecA + sec²A)
= (sin²A + cos²A) + 2(1) + 2(1) + cosec²A + sec²A
= 1 + 4 + (1 + cot²A) + (1 + tan²A)
= 7 + tan²A + cot²A = RHS. (Proved)

(ix) (cosec A − sin A)(sec A − cos A) = 1tan A + cot A
LHS = (1sin A – sin A)(1cos A – cos A) = (1 – sin²Asin A)(1 – cos²Acos A)
= (cos²Asin A)(sin²Acos A) = sin A cos A.
RHS = 1sin A/cos A + cos A/sin A = 1(sin²A + cos²A)/(sin A cos A) = 11/(sin A cos A) = sin A cos A.
Since LHS = RHS, the identity is proved.

(x) (1 + tan²A1 + cot²A) = (1 − tan A1 − cot A)² = tan²A
First part: 1 + tan²A1 + cot²A = sec²Acosec²A = tan²A.
Second part: (1 − tan A1 − 1/tan A)² = (1 − tan A(tan A – 1)/tan A)²
= (-(tan A – 1)tan Atan A – 1)² = (-tan A)² = tan²A.
All parts are equal to tan²A. (Proved)