MHT CET · Maths · Inverse Trigonometric Functions
With usual notations in \(\Delta \mathrm{ABC}\), if \(C=90^{\circ}\), then \(\tan ^{-1}\left(\frac{a}{b+c}\right)+\tan ^{-1}\left(\frac{b}{c+a}\right)=\)
- A \(\frac{\pi}{4}\)
- B \(\frac{\pi}{6}\)
- C \(\pi\)
- D \(\frac{\pi}{3}\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
\(\therefore \quad \tan ^{-1}\left(\frac{a}{b+c}\right)+\tan ^{-1}\left(\frac{b}{c+a}\right)\)
\(=\tan ^{-1}\left[\frac{\frac{a}{b+c}+\frac{b}{c+a}}{1-\left(\frac{a}{b+c}\right)\left(\frac{b}{c+a}\right)}\right]\)
\(=\tan ^{-1}\left[\frac{a c+a^{2}+b c+b^{2}}{a c+b c+a b+c^{2}-a b}\right]=\tan ^{-1}\left(\frac{a c+b c+c^{2}}{a c+b c+c^{2}}\right)\) \(\ldots\left[\because a^{2}+b^{2}=c^{2}\right.\), refer diagram \(]\)
\(=\tan ^{-1}(1)=\frac{\pi}{4}\)
\(=\tan ^{-1}\left[\frac{\frac{a}{b+c}+\frac{b}{c+a}}{1-\left(\frac{a}{b+c}\right)\left(\frac{b}{c+a}\right)}\right]\)
\(=\tan ^{-1}\left[\frac{a c+a^{2}+b c+b^{2}}{a c+b c+a b+c^{2}-a b}\right]=\tan ^{-1}\left(\frac{a c+b c+c^{2}}{a c+b c+c^{2}}\right)\) \(\ldots\left[\because a^{2}+b^{2}=c^{2}\right.\), refer diagram \(]\)
\(=\tan ^{-1}(1)=\frac{\pi}{4}\)
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