MHT CET · Maths · Trigonometric Ratios & Identities
\(\sec 2 \theta-\tan 2 \theta=\)
- A \(\tan \left(\frac{\pi}{4}-\theta\right)\)
- B \(\tan 2 \theta\)
- C \(\cot 2 \theta\)
- D \(\cot \left(\frac{\pi}{4}-\theta\right)\)
Answer & Solution
Correct Answer
(A) \(\tan \left(\frac{\pi}{4}-\theta\right)\)
Step-by-step Solution
Detailed explanation
\(\sec 2 \theta-\tan 2 \theta =\frac{1}{\cos 2 \theta}-\frac{\sin 2 \theta}{\cos 2 \theta}=\frac{1-\sin 2 \theta}{\cos 2 \theta} \)
\( =\frac{(\cos \theta-\sin \theta)^{2}}{\cos ^{2} \theta-\sin ^{2} \theta}=\frac{(\cos \theta-\sin \theta)^{2}}{(\cos \theta-\sin \theta)(\cos \theta+\sin \theta)}\)
\(=\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}=\frac{1-\tan \theta}{1+\tan \theta}=\frac{\tan \frac{\pi}{4}-\tan \theta}{1+\tan \frac{\pi}{4} \tan \theta}\)
\(=\tan \left(\frac{\pi}{4}-\theta\right)\)
\( =\frac{(\cos \theta-\sin \theta)^{2}}{\cos ^{2} \theta-\sin ^{2} \theta}=\frac{(\cos \theta-\sin \theta)^{2}}{(\cos \theta-\sin \theta)(\cos \theta+\sin \theta)}\)
\(=\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}=\frac{1-\tan \theta}{1+\tan \theta}=\frac{\tan \frac{\pi}{4}-\tan \theta}{1+\tan \frac{\pi}{4} \tan \theta}\)
\(=\tan \left(\frac{\pi}{4}-\theta\right)\)
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