Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
Difficulty select karke sirf is topic ke focused MCQs practice karein. Score, timer aur explanations student-friendly flow me milenge.
Showing 191-200 of 346 questions.
What is the simplified value of \(\sin^4 x-\cos^4 x\)?
C. \(\sin^2 x-\cos^2 x\)
इसे (\(\sin^2 x-\cos^2 x\)\(\sin^2 x+\cos^2 x\)) लिखें। दूसरा गुणनखंड (1) है। / Write it as (\(\sin^2 x-\cos^2 x\)\(\sin^2 x+\cos^2 x\)). The second factor is (1).
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What is the simplified value of \(\frac{1+\tan^2 x}{1+\cot^2 x}\)?
A. \(\tan^2 x\)
ऊपर \(1+\tan^2 x=\sec^2 x\) और नीचे \(1+\cot^2 x=\cosec^2 x\) है। अनुपात \(\frac{\sec^2 x}{\cosec^2 x}=\tan^2 x\) होता है। / The numerator is \(1+\tan^2 x=\sec^2 x\) and the denominator is \(1+\cot^2 x=\cosec^2 x\). Their ratio is \(\frac{\sec^2 x}{\cosec^2 x}=\tan^2 x\).
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How can \(\frac{1+\sin x}{1-\sin x}\) be written in terms of \(\sec x\) and \(\tan x\)?
B. (\(\sec x+\tan x\)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। / Rationalising the denominator gives (\frac{\(1+\sin x\)2}{\cos-2 x}). This equals (\(\sec x+\tan x\)2).
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What is \(\frac{1-\cos x}{1+\cos x}\) equal to?
C. \(\tan^2 \frac{x}{2}\)
अर्ध कोण पहचान के अनुसार \(\tan^2 \frac{x}{2}=\frac{1-\cos x}{1+\cos x}\)। अर्ध कोण रूपों को अलग से याद रखें। / By the half-angle identity, \(\tan^2 \frac{x}{2}=\frac{1-\cos x}{1+\cos x}\). Remember half-angle forms separately.
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What is \(\frac{\sin x}{1+\cos x}\) equal to?
B. \(\tan \frac{x}{2}\)
मानक अर्ध कोण पहचान \(\tan \frac{x}{2}=\frac{\sin x}{1+\cos x}\) है। ऐसे रूप में अर्ध कोण तुरंत पहचानें। / The standard half-angle identity is \(\tan \frac{x}{2}=\frac{\sin x}{1+\cos x}\). Identify the half-angle form quickly.
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If \(\sin x+\cos x=\frac{3}{2}\), what is the value of (\(\sin x-\cos x\)2)?
B. \(\frac{7}{4}\)
पहले (\(\sin x+\cos x\)2=1+2\sin x\cos x) से \(\sin x\cos x=\frac{5}{8}\) मिलता है। फिर (\(\sin x-\cos x\)2=1-2\sin x\cos x=\frac{7}{4})। / First, (\(\sin x+\cos x\)2=1+2\sin x\cos x) gives \(\sin x\cos x=\frac{5}{8}\). Then (\(\sin x-\cos x\)2=1-2\sin x\cos x=\frac{7}{4}).
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If \(\sin x-\cos x=\frac{1}{2}\), what is the value of \(\sin x\cos x\)?
A. \(\frac{3}{8}\)
(\(\sin x-\cos x\)2=1-2\sin x\cos x) लगाएँ। इससे \(\sin x\cos x=\frac{3}{8}\) मिलता है। / Use (\(\sin x-\cos x\)2=1-2\sin x\cos x). This gives \(\sin x\cos x=\frac{3}{8}\).
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If \(\sin x+\cos x=\frac{6}{5}\), what is the value of the square of \(\sin x-\cos x\)?
C. \(\frac{14}{25}\)
पहचान (\(\sin x+\cos x\)2+\(\sin x-\cos x\)2=2) का उपयोग करें। इसलिए मान \(2-\frac{36}{25}=\frac{14}{25}\) है। / Use the identity (\(\sin x+\cos x\)2+\(\sin x-\cos x\)2=2). Hence the value is \(2-\frac{36}{25}=\frac{14}{25}\).
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If \(\tan x+\cot x=5\), what is the value of \(\tan^2 x+\cot^2 x\)?
B. (23)
(\(\tan x+\cot x\)2=\tan-2 x+\cot-2 x+2) होता है। इसलिए मान (25-2=23) है। / (\(\tan x+\cot x\)2=\tan-2 x+\cot-2 x+2). Therefore, the value is (25-2=23).
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If \(\sec x-\tan x=\frac{1}{4}\), what is the value of \(\sec x+\tan x\)?
C. (4)
क्योंकि (\(\sec x-\tan x\)\(\sec x+\tan x\)=1) होता है। इसलिए दूसरा गुणनखंड (4) होगा। / Since (\(\sec x-\tan x\)\(\sec x+\tan x\)=1). Hence the other factor is (4).
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