Question

Using the trigonometric identity for the tangent of sum of two angles, and the tan of the difference of ${{180}^{\circ }}$and an angle, prove that:
$\dfrac{\tan {{69}^{\circ }}+\tan {{66}^{\circ }}}{1-\tan {{69}^{\circ }}\tan {{66}^{\circ }}}=-1$.

Answer 1

Hint: In this given question, we can use the transformation formula for \[tan\left( A+B \right)\] which is $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$. Here, we may use ${{69}^{\circ }}$in place of A and ${{66}^{\circ }}$in place of B and simplify the Left Hand Side (LHS) to $\tan ({{135}^{\circ }})$. Now, we can write ${{135}^{\circ }}$ as the sum of ${{180}^{\circ }}$ and $-{{45}^{\circ }}$. Then, reusing the transformation formula we can prove that the Left Hand Side (LHS) is equal to the Right Hand Side (RHS) of the given equation.

Complete step-by-step answer:
In the given question, we are asked to prove that $\dfrac{\tan {{69}^{\circ }}+\tan {{66}^{\circ }}}{1-\tan {{69}^{\circ }}\tan {{66}^{\circ }}}=-1$.
In order to do so and have a solution to this question, here, we are going to use the transformation formula of \[tan\left( A+B \right)\] that is $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$….…(1.1)
 where A will be substituted with ${{69}^{\circ }}$ and B will be substituted with ${{66}^{\circ }}$.
Now, as we can see that the LHS of the equation$\dfrac{\tan {{69}^{\circ }}+\tan {{66}^{\circ }}}{1-\tan {{69}^{\circ }}\tan {{66}^{\circ }}}=-1$ is in the form of $\dfrac{\tan A+\tan B}{1-\tan A\tan B}$, we can write is as equal to the form of \[tan\left( A+B \right)\] where A is equal to ${{69}^{\circ }}$ and B is equal to ${{66}^{\circ }}$, that is $\tan \left( {{69}^{\circ }}+{{66}^{\circ }} \right)=\tan \left( {{135}^{\circ }} \right)$.
Now, again we can write $\tan \left( {{135}^{\circ }} \right)$ as $\tan \left( {{180}^{\circ }}+(-{{45}^{\circ }}) \right)$.
Using equation 1.1, we can write $\tan \left( {{180}^{\circ }}+(-{{45}^{\circ }}) \right)$ as
$LHS=\tan \left( {{180}^{\circ }}+(-{{45}^{\circ }}) \right)$
$=\tan \left( {{180}^{\circ }}-{{45}^{\circ }} \right)$
$=-\tan \left( {{45}^{\circ }} \right)$ (as $\tan \left( {{180}^{\circ }}-\theta \right)=-\tan \theta $)
$=-1$ (as the value of $\tan {{45}^{\circ }}=1$)
$=RHS$
Hence, we satisfy the condition that the LHS is equal to the RHS.
Therefore, it has been proven that $\dfrac{\tan {{69}^{\circ }}+\tan {{66}^{\circ }}}{1-\tan {{69}^{\circ }}\tan {{66}^{\circ }}}=-1$.

Note: In this question, we may also have used ${{90}^{\circ }}\text{ }and\text{ }{{45}^{\circ }}$in place of ${{180}^{\circ }}\text{ }and\text{ -4}{{\text{5}}^{\circ }}$. But we did not use them because the value of $\tan {{90}^{\circ }}$ is not defined and we must not have got the required proof in search of which we started to solve this given question.