JEE Mains · Maths · STD 11 - 8. sequence and series
If \(\frac{tan(A-B)}{tan~A}+\frac{sin^{2}C}{sin^{2}A}=1,\) \(A, B, C\in(0,\frac{\pi}{2})\), then
- A tan A, tan C, tan B are in G.P.
- B tan A, tan B, tan C are in G.P.
- C tan A, tan C, tan B are in A.P.
- D tan A, tan B, tan C are in A.P.
Answer & Solution
Correct Answer
(A) tan A, tan C, tan B are in G.P.
Step-by-step Solution
Detailed explanation
\(\frac{tan~A-tan~B}{(1+tan~A~tan~B)tan~A}+\frac{1+cot^{2}A}{1+cot^{2}C}=1\) Put \(tan~A=x, tan~B=y, tan~C=z\) \(\therefore\)\(\frac{x-y}{(1+xy)x}+\frac{(x^{2}+1)z^{2}}{x^{2}(z^{2}+1)}=1\) \(\therefore\) \(x(x-y)(z^{2}+1)+z^{2}(1+x^{2})(1+xy)=(1+xy)x^{2}(1+z^{2})\) after solving…
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