Question

Simplify and find the value of the following \[\dfrac{3\tan {{25}^{\circ }}\tan {{40}^{\circ }}\tan {{50}^{\circ }}\tan {{65}^{\circ }}-\dfrac{1}{2}{{\tan }^{2}}{{60}^{\circ }}}{4\left( {{\cos }^{2}}{{29}^{\circ }}+{{\cos }^{2}}{{61}^{\circ }} \right)}\]

Answer 1

Hint: To solve the question we need to know the concept of trigonometric function and complementary angle. The angle here will be changed into the complementary so that the expression becomes simpler. We need to apply to the formulas for doing this.

Complete step by step solution:
The question ask us to find the value for the expression given to us which is \[\dfrac{3\tan {{25}^{\circ }}\tan {{40}^{\circ }}\tan {{50}^{\circ }}\tan {{65}^{\circ }}-\dfrac{1}{2}{{\tan }^{2}}{{60}^{\circ }}}{4\left( {{\cos }^{2}}{{29}^{\circ }}+{{\cos }^{2}}{{61}^{\circ }} \right)}\] . On analysing the expression we see that the tan of the angles present here cannot be solved without a calculator. Though we can see the relation between the angles given in the expression. So in the calculation check which all angles sum up to form ${{90}^{\circ }}$ . Starting with the numerator of the expression, we see that angle ${{25}^{\circ }}$and ${{65}^{\circ }}$ gives sum as ${{90}^{\circ }}$. Similarly the angles \[{{50}^{\circ }}\] and \[{{40}^{\circ }}\] add up giving ${{90}^{\circ }}$. Now we can write the angle as the difference of the complementary angle. So the expression becomes:
$\Rightarrow \dfrac{3\tan ({{90}^{\circ }}-{{65}^{\circ }})\tan ({{90}^{\circ }}-{{50}^{\circ }})\tan {{50}^{\circ }}\tan {{65}^{\circ }}-\dfrac{1}{2}{{\tan }^{2}}{{60}^{\circ }}}{4\left( {{\cos }^{2}}{{29}^{\circ }}+{{\cos }^{2}}{{61}^{\circ }} \right)}$
We know that we have a formula which says $\cot \alpha =\tan \left( {{90}^{\circ }}-\alpha \right)$ . On applying the same in the above expression we get:
$\Rightarrow \dfrac{3\cot {{65}^{\circ }}\cot {{50}^{\circ }}\tan {{50}^{\circ }}\tan {{65}^{\circ }}-\dfrac{1}{2}{{\tan }^{2}}{{60}^{\circ }}}{4\left( {{\cos }^{2}}{{29}^{\circ }}+{{\cos }^{2}}{{61}^{\circ }} \right)}$
The product of $\cot \alpha $and $\tan \alpha $ gives value as $1$, $\cot \alpha \times \tan \alpha =1$ .
So on operating the above expression with the above mentioned rule we get:
$\Rightarrow \dfrac{3\cot {{65}^{\circ }}\tan {{65}^{\circ }}\cot {{50}^{\circ }}\tan {{50}^{\circ }}-\dfrac{1}{2}{{\tan }^{2}}{{60}^{\circ }}}{4\left( {{\cos }^{2}}{{29}^{\circ }}+{{\cos }^{2}}{{61}^{\circ }} \right)}$
$\Rightarrow \dfrac{3-\dfrac{1}{2}{{\tan }^{2}}{{60}^{\circ }}}{4\left( {{\cos }^{2}}{{29}^{\circ }}+{{\cos }^{2}}{{61}^{\circ }} \right)}$
Since, we know that the value of $\tan {{60}^{\circ }}$ is $\sqrt{3}$. So on substituting the value we get:
$\Rightarrow \dfrac{3-\dfrac{1}{2}{{\left( \sqrt{3} \right)}^{2}}}{4\left( {{\cos }^{2}}{{29}^{\circ }}+{{\cos }^{2}}{{61}^{\circ }} \right)}$
$\Rightarrow \dfrac{3-\dfrac{3}{2}}{4\left( {{\cos }^{2}}{{29}^{\circ }}+{{\cos }^{2}}{{61}^{\circ }} \right)}$
On calculation the numerator becomes:
$\Rightarrow \dfrac{3}{2}\times \dfrac{1}{4\left( {{\cos }^{2}}{{29}^{\circ }}+{{\cos }^{2}}{{61}^{\circ }} \right)}$
Now we will be solving for the denominator. In the denominator as well we see the angle sum to ${{90}^{\circ }}$ . So it will be written as:
 $\Rightarrow \dfrac{3}{2}\times \dfrac{1}{4\left( {{\cos }^{2}}\left( {{90}^{\circ }}-{{61}^{\circ }} \right)+{{\cos }^{2}}{{61}^{\circ }} \right)}$
The formula used in the above expression will be $\sin \alpha =\cos \left( {{90}^{\circ }}-\alpha \right)$ and ${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$ . On applying these formula we get:
$\Rightarrow \dfrac{3}{2}\times \dfrac{1}{4\left( {{\sin }^{2}}{{61}^{\circ }}+{{\cos }^{2}}{{61}^{\circ }} \right)}$
$\Rightarrow \dfrac{3}{2}\times \dfrac{1}{4\times 1}$
$\Rightarrow \dfrac{3}{8}$
$\therefore $ The expression \[\dfrac{3\tan {{25}^{\circ }}\tan {{40}^{\circ }}\tan {{50}^{\circ }}\tan {{65}^{\circ }}-\dfrac{1}{2}{{\tan }^{2}}{{60}^{\circ }}}{4\left( {{\cos }^{2}}{{29}^{\circ }}+{{\cos }^{2}}{{61}^{\circ }} \right)}\] results to $\dfrac{3}{8}$.

Note: To solve the question we need to remember all the trigonometric formulas. The trigonometric function changes at the time of need but the sign depends upon the value of the angle. For example:
$\tan \left( {{180}^{\circ }}-\alpha \right)=-\tan \alpha $negative sign is applied here because in the second quadrant the trigonometric function $\tan $ is negative.