double-angle se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.
6 questions tagged with double-angle.
What is \(\cos^2 x-\sin^2 x\) equal to?
C. \(\cos 2x\)
दुगुना कोण पहचान \(\cos 2x=\cos^2 x-\sin^2 x\) है। परीक्षा में \(\cos 2x\) के तीनों रूप याद रखें। / The double angle identity is \(\cos 2x=\cos^2 x-\sin^2 x\). In exams, remember all three forms of \(\cos 2x\).
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What is \(2\sin x\cos x\) equal to?
B. \(\sin 2x\)
दुगुना कोण सूत्र के अनुसार \(2\sin x\cos x=\sin 2x\)। परीक्षा में यह identity बार-बार काम आती है। / By the double angle formula, \(2\sin x\cos x=\sin 2x\). This identity is used frequently in exams.
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What is \(\frac{1-\cos 2x}{\sin 2x}\) equal to?
D. \(\tan x\)
\(1-\cos 2x=2\sin^2 x\) और \(\sin 2x=2\sin x\cos x\) लगाएं। सरलीकरण से \(\tan x\) मिलता है। / Use \(1-\cos 2x=2\sin^2 x\) and \(\sin 2x=2\sin x\cos x\). Simplification gives \(\tan x\).
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What is \(\frac{\sin 2x}{1+\cos 2x}\) equal to?
C. \(\tan x\)
\(\sin 2x=2\sin x\cos x\) और \(1+\cos 2x=2\cos^2 x\) रखें। परिणाम \(\tan x\) है। / Use \(\sin 2x=2\sin x\cos x\) and \(1+\cos 2x=2\cos^2 x\). The result is \(\tan x\).
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If \(\tan x=2\), what is the value of \(\frac{1-\tan^2 x}{1+\tan^2 x}\)?
A. \( -\frac{3}{5} \)
सीधा रखने पर \(\frac{1-4}{1+4}=-\frac{3}{5}\) मिलता है। यह \(\cos 2x\) का भी रूप है। / Direct substitution gives \(\frac{1-4}{1+4}=-\frac{3}{5}\). This is also the form of \(\cos 2x\).
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What is \(\sin^2 \theta-\cos^2 \theta\) equal to?
B. \(-\cos 2\theta\)
क्योंकि \(\cos 2\theta=\cos^2 \theta-\sin^2 \theta\), इसलिए दिया गया रूप \(-\cos 2\theta\) है। परीक्षा में double angle identities को उल्टा भी पढ़ें। / Because \( \cos 2\theta=\cos^2 \theta-\sin^2 \theta\), the given form is \(-\cos 2\theta\). In exams read double angle identities in reverse too.
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