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Relation between Inverses of Trigonometric Functions and Their Reciprocal Functions

If sin A + ...

Question

If $sin A + sin B = C$ , $cos A + cos B = D$ , then the value of $sin (A + B) =$

\frac{(C^{2} + D^{2})}{2} - 1

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\frac{C D}{C^{2} + D^{2}}

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\frac{C^{2} + D^{2}}{2 C D}

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\frac{2 C D}{C^{2} + D^{2}}

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Solution

The correct option is A $\frac{(C^{2} + D^{2})}{2} - 1$
$W e h a v e sin A + sin B = C, a n d cos A + cos B = D$

$N o w, {(sin A + sin B)}^{2} = {sin}^{2} A + 2 sin A sin B + {sin}^{2} B = C^{2} A n d, {(cos A + cos B)}^{2} = {cos}^{2} A + 2 cos A c o s B + c o s^{2} B = D^{2}$

$B y a d d i n g t h e m w e g e t 1 + 2 (sin A sin B + cos A c o s B) = C^{2} + D^{2}$

or, $sin (A + B) = \frac{C^{2} + D^{2}}{2} - 1$ .

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Similar questions

Factorise:

$a^{2} + b^{2} - c^{2} - d^{2} + 2 a b - 2 c d$

a^{2} + b^{2} - c^{2} - d^{2} + 2 a b - 2 c d

If a, b, c, d are in G.p., prove that :

(i) $(a^{2} + b^{2}), (b^{2} + c^{2}), (c^{2} + d)^{2}$ are in G.P.

(ii) $(a^{2} - b^{2}), (b^{2} - c^{2}), (c^{2} - d)^{2}$ are in G.P.

(iii) $\frac{1}{a^{2} + b^{2}}, \frac{1}{b^{2} + c^{2}}, \frac{1}{c^{2} + d^{2}}$ are in G.P.

(iv) $(a^{2} + b^{2} + c^{2}), (a b + b c + c d), (b^{2} + c^{2} + d^{2})$

a, b, c, d

(a - c)^{2} + (b - c)^{2} + (b - d)^{2} - (a - d)^{2}

a

b

c

d

{(a - c)}^{2} + {(b - c)}^{2} + {(b - d)}^{2} - {(a - d)}^{2}

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