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23. If \( b \cos \theta=a \), then \( \operatorname{cosec} \theta+\cot \theta=? \) |
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Answer» We have, bcosθ = a ⇒ cosθ = \(\frac{a}{b}\) ∴ sinθ = \(\sqrt{1-cos^2θ}\) = \(\sqrt{1-\frac{a^2}{b^2}}\) = \(\sqrt{\frac{b^2-a^2}{b}}\). ∴ cotθ = \(\frac{cosθ}{sinθ}\) = \(\frac{\frac{a}{b}}{\frac{\sqrt{b^2-a^2}}{b}}\) = \(\frac{a}{\sqrt{b^2-a^2}}\) And cosecθ = \(\frac{1}{sinθ}\) = \(\frac{b}{\sqrt{b^2-a^2}}\) ∴ cosecθ + cotθ = \(\frac{b}{\sqrt{b^2-a^2}}\) + \(\frac{a}{\sqrt{b^2-a^2}}\) = \(\frac{b+a}{\sqrt{b^2-a^2}}\). |
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