Class 10 Maths Chapter 8 Introduction to Trigonometry Practice Paper

TRIGONOMETRY Class 10 - Mathematics Time Allowed: 59 minutes Maximum Marks: 25 1 . In △ OP Q right angled at P, OP = 7 cm, OQ - PQ = 1 cm. Determine the values of sin Q and cos Q. [3] cot A −cos A cosecA −1 [2] 2 . Prove that: = cot A +cos A cosecA +1 2 ∘ 2 ∘ 2 ∘ 5 cos 60 +4 sec 30 − tan 45 3 . Evaluate: [2] 2 ∘ 2 ∘ sin 30 + cos 30 − − − − − 1+sin A 4 . Prove √ = sec A + tan A, where the angles involved are acute angles for which the expressions are [2] 1−sin A defined. 3 sin θ −2 sin θ 5 . Prove = tan θ , where the angles involved are acute angles for which the expressions are defined. [2] 2 2 cos θ −cos θ 1 6 . Prove (cosec A - sin A) (sec A - cos A) = , where the angles involved are acute angles for which the [2] tan A +cot A expressions are defined.

2 1+sec A sin A 7 . Prove = , where the angles involved are acute angles for which the expressions are defined. [2] sec A 1−cos A 1−cos θ 2 8 . Prove ( cosecθ − cot θ ) = , where the angles involved are acute angles for which the expressions are [2] 1+cos θ defined. cos A −sin A +1 2 2 [2] 9 . Prove = cosec A + cot A, using the identity cosec A = 1 + cot A. where the angles involved are cos A +sin A −1 acute angles for which the expressions are defined. 1+sin A cos A 10 . Prove + = 2 sec A, where the angles involved are acute angles for which the expressions are [2] 1+sin A cos A defined.

2 2 2 2 [2] 11 . Prove (sin A + cosec A) + (cos A + sec A) = 7 + tan A + cot A, where the angles involved are acute angles for which the expressions are defined. tan A cot A 12 . Prove + = 1 + sec A cos ecA, where the angles involved are acute angles for which the [2] 1−cot A 1−tan A expressions are defined.

Solution

TRIGONOMETRY Class 10 - Mathematics 1. In △ OP Q , by Pythagoras theorem 2 2 2 OQ = OP + PQ 2 2 2 ⇒ ( P Q + 1 ) = O P + P Q [ ∵ OQ − P Q = 1 ⇒ OQ = 1 + P Q ] 2 2 2 ⇒ P Q + 2 P Q + 1 = 7 + P Q ⇒ 2 P Q + 1 = 49 ⇒ 2 P Q = 48 ⇒ P Q = 24 cm ∴ OQ − P Q = 1cm ⇒ OQ - 24 = 1 ⇒ OQ = 25 cm OP 7 Now, sin Q = = OQ 25 PQ 24 and, cos Q = = OQ 25 cot A −cos A

2. LHS = cot A +cos A cos A −cos A sin A = cos A +cos A sin A cos A −sin A cos A sin A = cos A +sin A cos A sin A cos A (1−sin A ) = cos A (1+sin A ) 1−sin A = 1+sin A 1 −1 sin A = 1 +1 sin A cosecA −1 = = RHS cosecA +1 2 ∘ 2 ∘ 2 ∘ 5 cos 60 +4 sec 30 − tan 45

3. Given: 2 ∘ 2 ∘ sin 30 + cos 30 2 2 1 2 2 5 ( ) +4 ( ) −(1 ) 2 √ 3 = 2 2 1 √ 3 ( ) + ( ) 2 2 1 4 5× +4× −1 4 3 = 1 3 + 4 4 5 16 + −1 4 3 = 1+3 4 15+64−12 12 = 4 4 67 12 = 1 67 = 12 − − − − − 1+sin A

4. L.H.S. √ 1−sin A − − − − − − − − − − 1+sin A 1+sin A = √ × √ 1−sin A 1+sin A − − − − − − − 2 (1+sin A ) 2 2 = √ [ ∵ ( a + b )( a − b ) = a − b ] 2 1− sin A − − − − − − − 2 (1+sin A ) 2 2 = √ [ ∵ 1 − sin θ = cos θ ] cos 2 A 1+sin A 1 sin A = = + = sec A + tan A = R ⋅ H . S . cos A cos A cos A

5. LHS 2 3 sin θ ( 1−2 sin θ ) sin θ −2 sin θ = = 2 2 2 cos θ −cos θ cos θ (2 cos θ −1) 2 2 2 sin θ ( cos θ + sin θ −2 sin θ ) 2 2 = ∵ cos θ + sin θ = 1 2 2 2 cos θ ( 2 cos θ − cos θ − sin θ ) 2 2 sin θ ( cos θ − sin θ ) = = tan θ 2 2 cos θ ( cos θ − sin θ ) = RHS

6. LHS = ( cosA − sin A )(sec A − cos A ) 2 2 1 1 1− sin A 1− cos A = ( − sin A ) ( − cos A ) = sin A cos A sin A cos A sin A cos A 2 2 cos A sin A 2 2 sin A cos A = , … … ∵ sin A + cos A = 1 = 2 2 sin A cos A sin A cos A + sin A cos A sin A cos A ........ Dividing the numerator and denominator by sin A cos A 1 = tan A +cot A = RHS 1 1+ 1+sec A cos A

7. LHS = = sec A 1 cos A cos A +1 cos A = = cos A + 1 = 1 + cos A 1 cos A 2 (1+cos A )(1−cos A ) 1− cos A = = 1−cos A 1−cos A 2 sin A 2 2 = ⋅ ∵ sin A + cos A = 1 1−cos A = RHS 2

8. LHS = ( cosec θ − cot θ ) 2 2 2 1 cos θ 1−cos θ (1−cos θ ) = ( − ) = ( ) = 2 sin θ sin θ sin θ sin θ 2 2 (1−cos θ ) (1−cos θ ) = = 2 1− cos θ (1−cos θ )(1+cos θ ) 1−cos θ = = RHS 1+cos θ

9. Taking L.H.S cos A −sin A +1 cos A +sin A −1 Dividing Numerator and Denominator by sin A cos A −sin A +1 sin A = cos A +sin A −1 sin A cos A sin A 1 − + sin A sin A sin A = cos A sin A 1 + − sin A sin A sin A cot A −1+ cosecA cos θ Using the formula cot θ = = sin θ cot A +1− cosecA 2 2 Using the identity cosec A = 1 + cot A 2 2 cot A − ( cose c A − cot A ) + cosecA = cot A +1− cosecA 2 2 (cot A + cosecA )− ( cose c A − cot A ) = cot A +1− cosecA (cot A + cosecA )(1− cosecA +cot A ) = cot A +1− cosecA = cot A + cosec A = R.H.S cos A 1+sin A

10. LHS = + 1+sin A cos A 2 2 cos A +(1+sin A ) 2 2 cos A +1+ sin A +2 sin A = = (1+sin A ) cos A (1+sin A ) cos A 1+1+2 sin A 2 2 = ∵ sin A + cos A = 1 (1+sin A ) cos A 2(1+sin A ) 2+2 sin A = = (1+sin A ) cos A (1+sin A ) cos A 2 1 = = 2 ⋅ = 2 sec A = RHS cos A cos A 2 2 2 2

11. To prove: (sinA + cosecA ) + (cosA + secA ) = 7 + tan A + cot A taking L.H.S 2 2 2 Using the formula (a + b ) = a + b + 2ab to get, 2 2 2 2 = (sin A + cosec A + 2sinA cosecA ) + (cos A + sec A + 2 cos A sec A ) 1 1 Since sin θ = and cos θ = cosecθ sec θ 2 2 1 2 2 1 = ( sin A + csc A + 2 sin A ) + ( cos A + sec A + 2 cos A ) sin A cos A 2 2 2 2 = sin A + cosec A + 2 + cos A + sec A + 2 2 2 2 2 = (sin A + cos A ) + cosec A + sec A + 2 + 2 2 2 2 2 2 2 Using the identities sin A + cos A = 1, sec A = 1 + tan A and cosec A = 1 + cot A to get 2 2 = 1 + 1 + tan A + 1 + cot A + 2 + 2 2 2 = 1 + 2 + 2 + 2 + tan A + cot A 2 2 = 7 + tan A + cot A = R.H.S.

Hence proved

12. LHS - tan A cot A + 1−cot A 1−tan A 1 tan A tan A = + 1 1−tan A 1− tan A tan A 1 = + tan A −1 tan A (1−tan A ) tan A 2 tan A 1 = + tan A −1 tan A (1−tan A ) 3 tan A −1 = tan A (tan A −1) 2 (tan A −1)( tan A +tan A +1) 3 3 2 2 = [a - b = (a - b ) (a + ab + b ) ] tan A (tan A −1) 2 tan A +tan A +1 = tan A = tan A + 1 + cot A sin A cos A = + + 1 cos A sin A 2 2 sin A + cos A = + 1 sin A cos A 1 = + 1 sin A cos A = sec A cosec A + 1 = R.H.S