Question

The value of $3\frac{{\sin }^{4}t+{\cos }^{4}t-1}{{\sin }^{6}t+{\cos }^{6}t-1}$ is equal to

Answer 1

Consider the given trigonometric expression.

$\Rightarrow 3\left[\frac{{\sin }^{4}t+{\cos }^{4}t-1}{{\sin }^{6}t+{\cos }^{6}t-1}\right]$

We know that,

${\sin }^2\theta +{\cos }^2\theta =1$

Now,

${\sin }^{4}t+{\cos }^{4}t={({\sin }^{2}t+{\cos }^{2}t)}^2-2{\sin }^{2}t{\cos }^{2}t$

${\sin }^{4}t+{\cos }^{4}t=1-2{\sin }^{2}t{\cos }^{2}t$ ……. (1)

Again,

${\sin }^{6}t+{\cos }^{6}t=({\sin }^{2}t+{\cos }^{2}t)\{{\sin }^{4}t+{\cos }^{4}t-{\sin }^{2}t{\cos }^{2}t\}$

${\sin }^{6}t+{\cos }^{6}t=1\{{({\sin }^{2}t+{\cos }^{2}t)}^2-2{\sin }^{2}t{\cos }^{2}t-{\sin }^{2}t{\cos }^{2}t\}$

${\sin }^{6}t+{\cos }^{6}t=1-3{\sin }^{2}t{\cos }^{2}t$ …… (2)

Substitute values from equations (1) and (2) in the given expression.

$\Rightarrow 3\left[\frac{{\sin }^{4}t+{\cos }^{4}t-1}{{\sin }^{6}t+{\cos }^{6}t-1}\right]$

$\Rightarrow 3\left[\frac{1-2{\sin }^{2}t{\cos }^{2}t-1}{1-3{\sin }^{2}t{\cos }^{2}t-1}\right]$

$\Rightarrow 3\left[\frac{-2{\sin }^{2}t{\cos }^{2}t}{-3{\sin }^{2}t{\cos }^{2}t}\right]$

$\Rightarrow 3\times \left(\frac{2}{3}\right)$

$\Rightarrow 2$

Hence, this is the required result.