TS EAMCET · Maths · Trigonometric Ratios & Identities
If \(\alpha, \beta\) and \(\gamma\) are length of the altitudes of a \(\triangle A B C\) with area \(\Delta\), then \(\frac{\Delta^2}{R^2}\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}\right)\) is equal to
- A \(\sin ^2 A+\sin ^2 B+\sin ^2 C\)
- B \(\cos ^2 A+\cos ^2 B+\cos ^2 C\)
- C \(\tan ^2 A+\tan ^2 B+\tan ^2 C\)
- D \(\cot ^2 A+\cot ^2 B+\cot ^2 C\)
Answer & Solution
Correct Answer
(A) \(\sin ^2 A+\sin ^2 B+\sin ^2 C\)
Step-by-step Solution
Detailed explanation
\(\because\) Area of triangle \(=\frac{1}{2}\) base \(\times\) altitude \(\therefore \Delta=\frac{1}{2} a \alpha=\frac{1}{2} b \beta=\frac{1}{2} c \gamma\) \(\Rightarrow \alpha=\frac{2 \Delta}{a}, \beta=\frac{2 \Delta}{b}\) and \(\gamma=\frac{2 \Delta}{c}\)…
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