MHT CET · Maths · Trigonometric Ratios & Identities
If \(\mathrm{A}+\mathrm{B}=225^{\circ}\), then \(\frac{\cot \mathrm{A}}{1+\cot \mathrm{A}} \cdot \frac{\cot \mathrm{B}}{1+\cot \mathrm{B}}\), if it exists, is equal to
- A 0
- B 1
- C 2
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
\(\frac{\cot A}{1+\cot A} \cdot \frac{\cot B}{1+\cot B} \)
\( =\frac{1}{(1+\tan A)(1+\tan B)} \)
\( =\frac{1}{\tan A+\tan B+1+\tan A \tan B}\)
\(=\frac{1}{1-\tan A \tan B+1+\tan A \tan B} \)\(\quad \cdots[\because \tan (A+B)=\tan 225^{\circ} \)
\( \Rightarrow \tan A+\tan B=1-\tan A \tan B\)
\( =\frac{1}{2}\)
\( =\frac{1}{(1+\tan A)(1+\tan B)} \)
\( =\frac{1}{\tan A+\tan B+1+\tan A \tan B}\)
\(=\frac{1}{1-\tan A \tan B+1+\tan A \tan B} \)\(\quad \cdots[\because \tan (A+B)=\tan 225^{\circ} \)
\( \Rightarrow \tan A+\tan B=1-\tan A \tan B\)
\( =\frac{1}{2}\)
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