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Values of Trigonometric Ratios

Question 5 v ...

Question

Question 5 (v)
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(v) $\frac{(c o s A - s i n A + 1)}{(c o s A + s i n A - 1)} = c o s e c A + c o t A$ , using the identity $c o s e c^{2} A = 1 + c o t^{2} A$ .

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Solution

$(v) (c o s A - s i n A + 1) / (c o s A + s i n A - 1) = c o s e c A + c o t A L . H . S . = \frac{(c o s A - s i n A + 1)}{(c o s A + s i n A - 1)} Dividing Numerator and Denominator by sin A, = \frac{\frac{(c o s A - s i n A + 1)}{s i n A}}{\frac{(c o s A + s i n A - 1)}{s i n A}} = \frac{(c o t A - 1 + c o s e c A)}{(c o t A + 1 - c o s e c A)} = \frac{(c o t A - c o s e c^{2} A + c o t^{2} A + c o s e c A)}{(c o t A + 1 - c o s e c A)} (∵ c o s e c^{2} A - c o t^{2} A = 1) = \frac{[(c o t A + c o s e c A) - (c o s e c^{2} A - c o t^{2} A)]}{(c o t A + 1 - c o s e c A)} = \frac{[(c o t A + c o s e c A) - (c o s e c A + c o t A) (c o s e c A - c o t A)]}{(1 - c o s e c A + c o t A)} = \frac{(c o t A + c o s e c A) (1 - c o s e c A + c o t A)}{(1 - c o s e c A + c o t A)} = c o t A + c o s e c A = R . H . S .$

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