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**Q.1. Evaluate:****(iii) cos 48Â° âˆ’ sin 42Â°****(iv) cosec 31Â° âˆ’ sec 59Â°****Sol. **

[âˆµ cos (90Â° - Î¸) = sin Î¸]

**Q.2. Show that: ****(i) tan 48Â° tan 23Â° tan 42Â° tan 67Â° = 1****(ii) cos 38Â° cos 52Â° âˆ’ sin 38Â° sin 52Â° = 0 ****Sol. **(i) LHS = tan 48Â° tan 23Â° tan 42Â° tan 67Â°

= tan 48Â° tan 23Â° tan (90Â° - 48Â°) tan (90Â° - 23Â°)

= tan 48Â° tan 23Â° cot 48Â° cot 23Â°

(ii) LHS = cos 38Â° cos 52Â° - sin 38Â° sin 52Â°

= cos 38Â° cos (90Â° - 38Â°) - sin 38Â°sin (90Â° - 38Â°)

= cos 38Â° sin 38Â°- sin 38Â° cos 38Â° = 0 = RHS**Q.3. If tan 2A = cot (A âˆ’ 18Â°), where 2A is an acute angle, find the value of A.****Sol.** tan 2A = cot (A - 18Â°)

â‡’ cot (90Â° - 2A) = cot (A - 18Â°)

[âˆµ cot (90Â° - Î¸) = tan Î¸]

â‡’ 90Â° - 2A = A - 18Â°

â‡’

âˆ´ âˆ A = 36Â°**Q.4. If tan A = cot B, prove that A + B = 90Â°.****Sol.** tan A = cot B

â‡’ tan A = tan (90Â° - B)

[âˆµ tan (90Â° - Î¸) = cot Î¸]

â‡’ A = 90Â° - B

â‡’ A + B = 90Â° Proved**Q.5. If sec 4A = cosec (A âˆ’ 20Â°), where 4A is an acute angle, find the value of A.****Sol. **sec 4A = cosec (A - 20Â°)

â‡’ cosec (90Â° - 4A) = cosec (A - 20Â°) [cosec (90Â° - Î¸) = sec Î¸]

â‡’ 90Â° - 4A = A - 20Â°

â‡’

âˆ´ âˆ A = 22Âº**Q.6. If A, B and C are interior angles of a triangle ABC, then show that**

**Sol. **In Î”ABC, A + B + C = 180Â° â‡’ B + C = 180Â° - A**Q.7. Express sin 67Â° + cos 75Â° in terms of trigonometric ratios of angles between 0Â° and 45Â°.****Sol.** sin 67Â° + cos 75Â° = sin (90Â° - 23Â°) + cos (90Â° - 15Â°)

= cos 23Â° + sin 15Â°

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