Evaluating the Trigonometric Expression

The problem asks us to evaluate the following expression: $$ \left( \frac{\sin 23^\circ \cos 67^\circ+\cos 23^\circ \sin 67^\circ}{\text{cosec}^2 15^\circ- \tan^2 75^\circ} \right)^3 $$ We need to simplify the numerator and the denominator separately before raising the result to the power of 3.

Simplifying the Numerator

The numerator is in the form of the sine addition identity: $$ \sin(A+B) = \sin A \cos B + \cos A \sin B $$ In our case, let $A = 23^\circ$ and $B = 67^\circ$. So, the numerator becomes: $$ \sin 23^\circ \cos 67^\circ+\cos 23^\circ \sin 67^\circ = \sin(23^\circ + 67^\circ) $$ $$ = \sin(90^\circ) $$ We know that $\sin(90^\circ) = 1$. Therefore, the value of the numerator is 1.

Simplifying the Denominator

The denominator is $\text{cosec}^2 15^\circ- \tan^2 75^\circ$. We can use complementary angle identities to simplify this. Recall that: $$ \tan \theta = \cot(90^\circ - \theta) $$ Applying this to $\tan 75^\circ$: $$ \tan 75^\circ = \tan(90^\circ - 15^\circ) = \cot 15^\circ $$ Substituting this back into the denominator expression: $$ \text{cosec}^2 15^\circ- \tan^2 75^\circ = \text{cosec}^2 15^\circ - (\cot 15^\circ)^2 $$ $$ = \text{cosec}^2 15^\circ - \cot^2 15^\circ $$ Now, we use the Pythagorean identity involving cosecant and cotangent: $$ \text{cosec}^2 \theta - \cot^2 \theta = 1 $$ Using this identity with $\theta = 15^\circ$: $$ \text{cosec}^2 15^\circ - \cot^2 15^\circ = 1 $$ Thus, the value of the denominator is 1.

Calculating the Final Value

Now we substitute the simplified values of the numerator and denominator back into the original expression: $$ \left( \frac{\text{Numerator}}{\text{Denominator}} \right)^3 = \left( \frac{1}{1} \right)^3 $$ $$ = (1)^3 $$ $$ = 1 $$ The final evaluated value of the expression is 1.