MHT CET · Maths · Trigonometric Ratios & Identities
If \(\tan \theta+\sin \theta=a\) and \(\tan \theta-\sin \theta=b\), then the values of \(\cot \theta\) and \(\operatorname{cosec} \theta\) are respectively
- A \(\frac{1}{a+b}, \frac{1}{a-b}\)
- B \(\frac{2}{a+b}, \frac{2}{a-b}\)
- C \(\frac{2}{a-b}, \frac{2}{a+b}\)
- D \(\frac{1}{a-b}, \frac{1}{a+b}\)
Answer & Solution
Correct Answer
(B) \(\frac{2}{a+b}, \frac{2}{a-b}\)
Step-by-step Solution
Detailed explanation
We have
\(\tan \theta+\sin \theta=a\ldots(1)\) and
\(\tan \theta-\sin \theta=b\ldots(2)\)
Adding equation (1) \& (2), we get
\(2 \tan \theta=a+b \Rightarrow \tan \theta=\frac{a+b}{2} \Rightarrow \cot \theta=\frac{2}{a+b}\)
By equation (1) - equation (2), we get
\(2 \sin \theta=a-b \quad \Rightarrow \sin \theta=\frac{a-b}{2} \Rightarrow \operatorname{cosec} \theta=\frac{2}{a-b}\)
\(\tan \theta+\sin \theta=a\ldots(1)\) and
\(\tan \theta-\sin \theta=b\ldots(2)\)
Adding equation (1) \& (2), we get
\(2 \tan \theta=a+b \Rightarrow \tan \theta=\frac{a+b}{2} \Rightarrow \cot \theta=\frac{2}{a+b}\)
By equation (1) - equation (2), we get
\(2 \sin \theta=a-b \quad \Rightarrow \sin \theta=\frac{a-b}{2} \Rightarrow \operatorname{cosec} \theta=\frac{2}{a-b}\)
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