Trig Simplification (2)

Given :
$\frac {\sec\alpha \sec\beta \sec\gamma}{\tan\alpha+\tan\beta+\tan\gamma-\tan\alpha\tan\beta\tan\gamma}$
Since, $\sec x= \frac{1}{\cos x}$ for any $\angle x$ ,
therefore, we can write the given expression as:
$\Rightarrow\frac {1} {\cos\alpha\cos\beta\cos\gamma(\tan\alpha+\tan\beta+\tan\gamma-\tan\alpha\tan\beta\tan\gamma)}$
Now, since we know that $\tan\alpha = \frac{\sin\alpha}{\cos\alpha}$ ,
we can write the above expression as:
$\Rightarrow\frac{1}{(\cos\alpha\cos\beta\cos\gamma)(\frac{\sin\alpha}{\cos\alpha}+\frac{\sin\beta}{\cos\beta}+\frac{\sin\gamma}{\cos\gamma}-\frac{\sin\alpha\sin\beta\sin\gamma}{\cos\alpha\cos\beta\cos\gamma})}$
$=\frac{1}{(\sin\alpha\cos\beta\cos\gamma)+(\sin\beta\cos\alpha\cos\gamma)+(\sin\gamma\cos\alpha\cos\beta)-(\sin\alpha\sin\beta\sin\gamma)}$
$=\frac{1}{{\sin\alpha(\cos\beta\cos\gamma-\sin\beta\sin\gamma)}+{\cos\alpha(\sin\beta\cos\gamma+\cos\beta\sin\gamma)}}$
$=\frac{1}{{\sin\alpha\cos(\beta+\gamma)}+{\cos\alpha\sin(\beta+\gamma)}}$
$=\frac{1}{\sin(\alpha+\beta+\gamma)}$
$=\mathrm{cosec}(\alpha+\beta+\gamma)$

Multiply both the denominator and numerator with cosa.cosb.cosy. Then you can take cosy as common from the first two terms and siny common from the last two terms of denominator. After some easy steps, you will get 1/sin (a+b+y) which equals cosec (a+b+y)

check for alpha=beta=gamma=45 its comes to be $\sqrt{ 2}$ checking for the options we get csc(45+45+45)=csc(135)= $\sqrt{ 2}$

Multiply both the denominator and numerator with cos alpha, cos beta, cos gamma. now use the following identities after taking common terms out 1) sin(a+b)=sin a x cos b + cos a x sin b 2)cos(a+b)=cos a x cos b - sin a x sin b