Question

The expression
$(\alpha \tan \gamma +\beta \cot \gamma )(\alpha \cot \gamma +\beta \tan \gamma )-4\alpha \beta {\cot }^{2}2\gamma$

• A. independent of $\alpha ,\beta$
• B. dependent of $\gamma ,\alpha$
• C. dependent on $\gamma$
• D. dependent on $\alpha ,\beta$

Answer 1

The correct option is A independent of $\alpha ,\beta$

$(\alpha \tan \gamma +\beta \cot \gamma )(\alpha \cot \gamma +\beta \tan \gamma )-4\alpha \beta {\cot }^{2}2\gamma$

$\Rightarrow {\alpha }^2+\alpha \beta {\tan }^2\gamma +\alpha \beta {\cot }^2\gamma +{\beta }^2-4\alpha \beta {\cot }^{2}2\gamma$

$\Rightarrow {\alpha }^2+\alpha \beta {\tan }^2\gamma +\alpha \beta {\cot }^2\gamma +{\beta }^2-4\alpha \beta \times \frac{1}{4}(\cot \gamma -\tan \gamma {)}^2$

As$[\cot 2\gamma =\frac{\cos 2\gamma }{\sin 2\gamma }=\frac{{\cos }^2\gamma -{\sin }^2\gamma }{2\sin \gamma \cos \gamma }=\frac{\cos \gamma \cos \gamma }{2\sin \gamma \cos \gamma }-\frac{\sin \gamma \sin \gamma }{2\sin \gamma \cos \gamma }=\frac{\cos \gamma }{2\sin \gamma }-\frac{\sin \gamma }{2\cos \gamma }=\frac{1}{2}(\cot \gamma -\tan \gamma )$]

$\Rightarrow {\alpha }^2+\alpha \beta {\tan }^2\gamma +\alpha \beta {\cot }^2\gamma +{\beta }^2-\alpha \beta ({\cot }^2\gamma +{\tan }^2\gamma -2)$.

$\Rightarrow {\alpha }^2+\alpha \beta {\tan }^2\gamma +\alpha \beta {\cot }^2\gamma +{\beta }^2-\alpha \beta {\cot }^2\gamma -\alpha \beta {\tan }^2\gamma +2\alpha \beta$

$\Rightarrow {\alpha }^2+{\beta }^2+2\alpha \beta$

$\Rightarrow (\alpha +\beta {)}^2$

$\therefore$ Dependent on $\alpha ,\beta$.