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**Illustration 1: If cos (a + b) = 4/5, sin (a-b) = 5/13 and a and b lie between 0 to Ï€/4, find tan 2a.****Solution: **It is given that cos (a + b) = 4/5, sin (a-b) = 5/13

It follows that sin (a + b) = 3/5, cos (a-b) = 12/13

Hence, tan (a+b) = 3/4 and tan (a-b) = 5/12

Hence, this implies

tan [(a+b) + a-b] =[tan (a+b) + tan (a-b)]/ [1 + tan(a+b)tan(a-b)]

= (3/4 + 5/12)/ (1 â€“ 3/4. 5/12)

Hence, tan 2a = 14/12. 48/33 = 56/33.

**Illustration 2: If A + B + C = 180 ^{0}, prove that **

So consider LHS = sin (B + C â€“ A) + sin (C + A â€“ B) + sin (A + B â€“ C)

= sin (Ï€â€“ A â€“ A) + sin (Ï€ â€“ B â€“ B) + sin (Ï€â€“ C â€“ C)(since A + B + C = Ï€)

= sin 2A + sin 2B + sin 2C

= 4 sin A sin B sin C

**Illustration 3: Solve the equation 5 sinÎ¸ â€“ 2 cos ^{2}Î¸ â€“ 1 = 0**

or,5 sinÎ¸ â€“ 2 (1 â€“ sin

or,2 sin

or,(sin Î¸ + 3) (2 sin Î¸ â€“ 1) = 0

âˆ´sin Î¸ = 3 or, sin Î¸ = 1/2

if sin Î¸ = 3

This is not possible as range of sine is [â€“1, 1]

If sin Î¸ = 1/2

or, sin Î¸ = sin Ï€/6

â‡’ Î¸ = n Ï€ + (â€“1)

where n = 0, Â±1, Â±2 â€¦â€¦

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