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Question Number 88867 by liki last updated on 13/Apr/20

Commented by liki last updated on 13/Apr/20

$. . h e l p m e p l e a s e q u e s t i o n r o m a n i, i i & i i$
	Answered by TANMAY PANACEA. last updated on 13/Apr/20

	$a c o s θ + b s i n θ = c$ $t = t a n \frac{θ}{2}$ $a (\frac{1 - t^{2}}{1 + t^{2}}) + b (\frac{2 t}{1 + t^{2}}) = c$ $a - {a t}^{2} + 2 b t = c + {c t}^{2}$ $t^{2} (a + c) + t (- 2 b) + (c - a) = 0$ $t_{1} + t_{2} = \frac{2 b}{a + c} a n d t_{1} t_{2} = \frac{c - a}{c + a}$ $t_{1} = t a n (\frac{α}{2}) t_{2} = t a n (\frac{β}{2})$ $s i n (α + β) = s i n 2 (\frac{α + β}{2}) = \frac{2 t a n (\frac{α + β}{2})}{1 + {t a n}^{2} (\frac{α + β}{2})}$ $n o w s i n (α + β) = \frac{2 \times \frac{t a n \frac{α}{2} + t a n \frac{β}{2}}{1 - t a n \frac{α}{2} . t a n \frac{β}{2}}}{1 + {(\frac{t a n \frac{α}{2} + t a n \frac{β}{2}}{1 - t a n \frac{α}{2} . t a n \frac{β}{2}})}^{2}}$ $= \frac{2 \times \frac{t_{1} + t_{2}}{1 - t_{1} t_{2}}}{1 + {(\frac{t_{1} + t_{2}}{1 - t_{1} t_{2}})}^{2}}$ $= \frac{2 (t_{1} + t_{2}) (1 - t_{1} t_{2})}{{(1 - t_{1} t_{2})}^{2} + {(t_{1} + t_{2})}^{2}}$ $= \frac{2 (\frac{2 b}{a + c}) (1 - \frac{c - a}{c + a})}{{(1 - \frac{c - a}{c + a})}^{2} + {(\frac{2 b}{c + a})}^{2}}$ $= \frac{\frac{4 b \times 2 a}{{(c + a)}^{2}}}{\frac{4 a^{2}}{{(c + a)}^{2}} + \frac{4 b^{2}}{{(c + a)}^{2}}} = \frac{8 a b}{4 (a^{2} + b^{2})} = \frac{2 a b}{a^{2} + b^{2}}$ $2) t a n (α + β) = t a n 2 (\frac{α + β}{2}) = \frac{2 t a n (\frac{α + β}{2})}{1 - {t a n}^{2} (\frac{α + β}{2})}$ $= \frac{2 (\frac{t_{1} + t_{2}}{1 - t_{1} t_{2}})}{1 - {(\frac{t_{1} + t_{2}}{1 - t_{1} t_{2}})}^{2}} = \frac{2 (t_{1} + t_{2}) (1 - t_{1} t_{2})}{{(1 - t_{1} t_{2})}^{2} - {(t_{1} + t_{2})}^{2}}$ $= \frac{2 (\frac{2 b}{a + c}) (1 - \frac{c - a}{c + a})}{{(1 - \frac{c - a}{c + a})}^{2} - {(\frac{2 b}{c + a})}^{2}}$ $= \frac{4 b \times 2 a}{4 a^{2} - 4 b^{2}} = \frac{2 a b}{a^{2} - b^{2}}$ $3) c o s (α - β) = \frac{1 - {t a n}^{2} (\frac{α - β}{2})}{1 + {t a n}^{2} (\frac{α - β}{2})}$ $= \frac{1 - {(\frac{t_{1} - t_{2}}{1 + t_{1} t_{2}})}^{2}}{1 + {(\frac{t_{1} - t_{2}}{1 + t_{1} t_{2}})}^{2}}$ $= \frac{{(1 + t_{1} t_{2})}^{2} - {(t_{1} - t_{2})}^{2}}{{(1 + t_{1} t_{2})}^{2} + {(t_{1} - t_{2})}^{2}}$ $= \frac{{(1 + t_{1} t_{2})}^{2} - {(t_{1} + t_{2})}^{2} + 4 t_{1} t_{2}}{{(1 + t_{1} t_{2})}^{2} + {(t_{1} + t_{2})}^{2} - 4 t_{1} t_{2}}$ $= \frac{{(1 + \frac{c - a}{c + a})}^{2} - {(\frac{2 b}{c + a})}^{2} + 4 . \frac{c - a}{c + a}}{{(1 + \frac{c - a}{c + a})}^{2} + {(\frac{2 b}{c + a})}^{2} - 4 . (\frac{c - a}{c + a})}$ $= \frac{\frac{4 c^{2} - 4 b^{2} + 4 (c^{2} - a^{2})}{{(c + a)}^{2}}}{\frac{4 c^{2} + 4 b^{2} - 4 (c^{2} - a^{2})}{{(c + a)}^{2}}}$ $= \frac{c^{2} - b^{2} + c^{2} - a^{2}}{c^{2} + b^{2} - c^{2} + a^{2}}$ $\frac{2 c^{2} - a^{2} - b^{2}}{a^{2} + b^{2}}$
	Commented by liki last updated on 13/Apr/20

	$. . . t h a n k y o u s o m u c h s i r b l e s s e d$
	Commented by TANMAY PANACEA. last updated on 13/Apr/20

	$m o s t w e l c o m e$
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