What is the simplified value of \(\frac{1-\cos^2 x}{\sin^2 x}\)?\(\frac{1-\cos^2 x}{\sin^2 x}\) का सरल मान क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \(1\)\(1\)
Using the Pythagorean identity \(\sin^2 x+\cos^2 x=1\), we get \(1-\cos^2 x=\sin^2 x\). Therefore, \(\frac{1-\cos^2 x}{\sin^2 x}=\frac{\sin^2 x}{\sin^2 x}=1\), provided \(\sin x\ne 0\). Neither \(\sin x\), \(\cos x\), nor \(\tan x\) is the simplified value of this ratio. Exam tip: first replace \(1-\cos^2 x\) with \(\sin^2 x\) in such questions.पाइथागोरस सर्वसमिका \(\sin^2 x+\cos^2 x=1\) से \(1-\cos^2 x=\sin^2 x\) प्राप्त होता है। अतः \(\frac{1-\cos^2 x}{\sin^2 x}=\frac{\sin^2 x}{\sin^2 x}=1\), जहाँ \(\sin x\ne 0\) हो। \(\sin x\), \(\cos x\) या \(\tan x\) स्वयं इस अनुपात के सरल मान नहीं हैं। परीक्षा सुझाव: ऐसे प्रश्नों में पहले \(1-\cos^2 x\) को \(\sin^2 x\) से बदलें।
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