Question

If $2ta{n}^2\alpha ta{n}^2\beta ta{n}^2\gamma +ta{n}^2\alpha ta{n}^2\beta +ta{n}^2\beta ta{n}^2\gamma +ta{n}^2\gamma ta{n}^2\alpha =1$ (where, $\alpha ,\beta ,\gamma \ne (2n+1)\frac{\pi }{2},\ne n\pi ,nϵI$) then

• A. $si{n}^2\alpha +si{n}^2\beta +si{n}^2\gamma =1$
• B. $co{s}^2\alpha +co{s}^2\beta +co{s}^2\gamma =1$
• C. $cose{c}^2\alpha cose{c}^2\beta +cose{c}^2\beta cose{c}^2\gamma +cose{c}^2\gamma cose{c}^2\alpha =cose{c}^2\alpha cose{c}^2\beta cose{c}^2\gamma$
• D. $co{s}^2\alpha +co{s}^2\beta +co{s}^2\gamma =2$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $si{n}^2\alpha +si{n}^2\beta +si{n}^2\gamma =1$
C $cose{c}^2\alpha cose{c}^2\beta +cose{c}^2\beta cose{c}^2\gamma +cose{c}^2\gamma cose{c}^2\alpha =cose{c}^2\alpha cose{c}^2\beta cose{c}^2\gamma$
D $co{s}^2\alpha +co{s}^2\beta +co{s}^2\gamma =2$

Dividing by $ta{n}^2\alpha ta{n}^2\beta ta{n}^2\gamma$, we get
$2+co{t}^2\gamma +co{t}^2\alpha +co{t}^2\beta =co{t}^2\alpha co{t}^2\beta co{t}^2\gamma$
$\Rightarrow -1+\sum cose{c}^2\alpha =\pi (cose{c}^2\alpha -1)$
$\Rightarrow -1+\sum cose{c}^2\alpha =-1+\sum cose{c}^2\alpha -\sum cose{c}^2\alpha .cose{c}^2\beta +\pi \left(cose{c}^2\alpha \right)$
$\Rightarrow \sum cose{c}^2\alpha .cose{c}^2\beta =\pi \left(cose{c}^2\alpha \right)$
$\Rightarrow si{n}^2\alpha +si{n}^2\beta +si{n}^2\gamma =1$.