Match the following. Given cos α + cos β = a and sin α + sin β = b . List-I List-II (a) cos ( α + β ) (p) a 2 + b 2 2 ab (b) sin ( α + β ) (q) a b (c) cos ( α − β ) (r) a 2 + b 2 a 2 − b 2 (d) t

Question

Match the following. Given cos α + cos β = a and sin α + sin β = b . List-I List-II (a) cos ( α + β ) (p) a 2 + b 2 2 ab ​ (b) sin ( α + β ) (q) a b ​ (c) cos ( α − β ) (r) a 2 + b 2 a 2 − b 2 ​ (d) tan 2 α + β ​ (s) 2 a 2 + b 2 − 2 ​ (t) ( 2 a 2 + b 2 ​ ) − 1 (One or more entries can match more than one option.)

TestHub · Accepted Answer

a = 2 cos 2 α + β ​ cos 2 α − β ​ , b = 2 sin 2 α + β ​ cos 2 α − β ​ , so tan 2 α + β ​ = a b ​ (d-q). cos ( α + β ) = 1 + a 2 b 2 ​ 1 − a 2 b 2 ​ ​ = a 2 + b 2 a 2 − b 2 ​ ( a − r ) ; sin ( α + β ) = 1 + a 2 b 2 ​ 2 ⋅ a b ​ ​ = a 2 + b 2 2 ab ​ ( b − p ) . a 2 + b 2 = 2 + 2 cos ( α − β ) , so cos ( α − β ) = 2 a 2 + b 2 − 2 ​ = 2 a 2 + b 2 ​ − 1 , which is both (s) and (t) (c-s,t).

List-I	List-II
(a) $cos (α + β)$	(p) $\frac{2 ab}{a ^{2} + b ^{2}}$
(b) $sin (α + β)$	(q) $\frac{b}{a}$
(c) $cos (α - β)$	(r) $\frac{a ^{2} - b ^{2}}{a ^{2} + b ^{2}}$
(d) $tan \frac{α + β}{2}$	(s) $\frac{a ^{2} + b ^{2} - 2}{2}$
	(t) $(\frac{a ^{2} + b ^{2}}{2}) - 1$

Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

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