Difficulty distribution
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Practice Trigonometry - Mathematics previous year questions organised from real papers, with year-wise coverage and clear topic navigation.
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Year-wise coverage for Trigonometry. Each bar uses a separate theme-derived color.
How the classified questions are distributed by difficulty.
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Explore previous-paper coverage, trends and focused practice for Trigonometric Ratios And Identities.
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Explore previous-paper coverage, trends and focused practice for Trigonometric Equations.
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| Paper | Year / session | Questions in this view | Open |
|---|---|---|---|
| TG EAPCET 2025 ONLINE 2ND MAY EVENING SHIFT | 2025 | 10 | View paper |
| TG EAPCET 2025 ONLINE 2ND MAY MORNING SHIFT | 2025 | 12 | View paper |
| TG EAPCET 2025 ONLINE 3RD MAY EVENING SHIFT | 2025 | 9 | View paper |
| TG EAPCET 2025 ONLINE 3RD MAY MORNING SHIFT | 2025 | 9 | View paper |
| TG EAPCET 2025 ONLINE 4TH MAY EVENING SHIFT | 2025 | 9 | View paper |
| TG EAPCET 2025 ONLINE 4TH MAY MORNING SHIFT | 2025 | 8 | View paper |
| TG EAPCET 2024 (Online) 9th May Morning Shift | 2024 | 8 | View paper |
| TG EAPCET 2024 ONLINE 10TH MAY EVENING SHIFT | 2024 | 10 | View paper |
| TG EAPCET 2024 ONLINE 10TH MAY MORNING SHIFT | 2024 | 12 | View paper |
| TG EAPCET 2024 ONLINE 11TH MAY MORNING SHIFT | 2024 | 8 | View paper |
| TG EAPCET 2024 ONLINE 9TH MAY EVENING SHIFT | 2024 | 8 | View paper |
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Consider the following statements
Assertion (A) For $x \in R-\{1\}$;
$$\frac{d}{d x}\left(\tan ^{-1}\left(\frac{1+x}{1-x}\right)\right)=\frac{d}{d x}\left(\tan ^{-1} x\right)$$
Reason (R) For $x<1, \tan ^{-1}\left(\frac{1+x}{1-x}\right)=\frac{\pi}{4}+\tan ^{-1} x$, for
$$x>1, \tan ^{-1}\left(\frac{1+x}{1-x}\right)=-\frac{3 \pi}{4}+\tan ^{-1} x$$
The correct answer is
The general solution of the equation $\sqrt{6-5 \cos x+7 \sin ^2 x}-\cos x=0$ also satisfies the equation