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How can (\frac{1+\sin x}{1-\sin x}) be written in terms of (\sec x) and (\tan x)?(\frac{1+\sin x}{1-\sin x}) को (\sec x) और (\tan x) के रूप में किसके बराबर लिखा जा सकता है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: ((\sec x+\tan x)^2)((\sec x+\tan x)^2)
Rationalising the denominator gives (\frac{(1+\sin x)^2}{\cos^2 x}). This equals ((\sec x+\tan x)^2).हर को परिमेय करने पर (\frac{(1+\sin x)^2}{\cos^2 x}) मिलता है। यह ((\sec x+\tan x)^2) के बराबर है।
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