Prove that-i 1-sin x-a sin x-b- x-a- x-b-sin a-bii 1-sin x-a cos x-b- x-a-tan x-b-cos a-b

=1/sin(a b)= 1/sin(a b) [sin(x b) (x a)/cos(x a)cos(x b) ] =1/sin(a b)= [sin(x b)cos(x a) cos(x b)sin(x a)/cos(x a)cos(x b)cos(x a)cos(x b) ] = 1/sin(a b) [ ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Compound Angles

Prove that:i ...

Question

Prove that:
(i) $\frac{1}{s i n (x - a) s i n (x - b)} = \frac{c o t (x - a) - c o t (x - b)}{s i n (a - b)}$
(ii) $\frac{1}{s i n (x - a) c o s (x - b)} = \frac{c o t (x - a) + t a n (x - b)}{c o s (a - b)}$

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Solution

= $\frac{1}{s i n (a - b)} = \frac{1}{s i n (a - b)} [\frac{s i n (x - b) - (x - a)}{c o s (x - a) c o s (x - b)}] = \frac{1}{s i n (a - b)} = [\frac{s i n (x - b) c o s (x - a) - c o s (x - b) s i n (x - a)}{c o s (x - a) c o s (x - b) c o s (x - a) c o s (x - b)}] = \frac{1}{s i n (a - b)} [\frac{s i n (x - b)}{c o s (x - b)} - \frac{s i n (x - a)}{c o s (x - a)}] = \frac{1}{s i n (a - b)} [t a n (x - b) - t a n (x - s)] = \frac{t a n (x - b) - t a n (x - a)}{s i n (a - b)}$
= RHS
$∴$ LHS = RHS Hence proved.

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