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JEE Main Mock Test Series 2020 & Previous Year Papers

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JEE : Notes | EduRev

The document Notes | EduRev is a part of the JEE Course JEE Main Mock Test Series 2020 & Previous Year Papers.
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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 = 1800, prove that sin (B + C â€“ A) + sin (C + A â€“ B) + sin (A + B â€“ C) = 4 sin A sin B sin C.
Solution:We have sin (B + C â€“ A) + sin (C + A â€“ B) + sin (A + B â€“ C) = 4 sin A sin B sin C
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 cos2Î¸ â€“ 1 = 0
Solution: Given, 5sinÎ¸ â€“ 2cos2Î¸ â€“ 1 = 0
or,5 sinÎ¸ â€“ 2 (1 â€“ sin2Î¸) â€“ 1 = 0
or,2 sin2 Î¸ + 5 sin Î¸ â€“ 3 = 0
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)n Ï€/6
where n = 0, Â±1, Â±2 â€¦â€¦

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