Question

Check if the given statement is correct or not: “The value of $\tan {{75}^{\circ }} = \dfrac{\sqrt{3}+1}{2\sqrt{2}}$.”
(a) True
(b) False

Answer 1

Hint: To check if the given statement is correct or not, use the trigonometric identity $\tan \left( x+y \right)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$. Substitute $x={{45}^{\circ }},y={{30}^{\circ }}$ in the formula and simplify the expression by substituting the values $\tan {{45}^{\circ }}=1,\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}$. Simplify the expression and calculate the exact value of $\tan {{75}^{\circ }}$.

Complete Step-by-step answer:
We have to check if the given statement “The value of $\tan {{75}^{\circ }}=\dfrac{\sqrt{3}+1}{2\sqrt{2}}$” is correct or not.
We will calculate the value of $\tan {{75}^{\circ }}$.
To do so, we will use the trigonometric identity $\tan \left( x+y \right)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$.
Substituting $x={{45}^{\circ }},y={{30}^{\circ }}$ in the above formula, we have $\tan \left( {{45}^{\circ }}+{{30}^{\circ }} \right)=\dfrac{\tan {{45}^{\circ }}+\tan {{30}^{\circ }}}{1-\tan {{45}^{\circ }}\tan {{30}^{\circ }}}$.
We know the trigonometric values $\tan {{45}^{\circ }}=1,\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}$. Substituting these values in the above equation, we have $\tan \left( {{45}^{\circ }}+{{30}^{\circ }} \right)=\dfrac{\tan {{45}^{\circ }}+\tan {{30}^{\circ }}}{1-\tan {{45}^{\circ }}\tan {{30}^{\circ }}}=\dfrac{1+\dfrac{1}{\sqrt{3}}}{1-1\left( \dfrac{1}{\sqrt{3}} \right)}$.
Simplifying the above expression, we have $\tan \left( {{45}^{\circ }}+{{30}^{\circ }} \right)=\dfrac{1+\dfrac{1}{\sqrt{3}}}{1-1\left( \dfrac{1}{\sqrt{3}} \right)}=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}$.
We will now rationalize the above equation. To do so, we will multiply the numerator and denominator by $\sqrt{3}+1$.
Thus, we have $\tan \left( {{75}^{\circ }} \right)=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\times \dfrac{\sqrt{3}+1}{\sqrt{3}+1}$.
We know the algebraic identities ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and $\left( x+y \right)\left( x-y \right)={{x}^{2}}-{{y}^{2}}$.
Thus, we have $\tan \left( {{75}^{\circ }} \right)=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\times \dfrac{\sqrt{3}+1}{\sqrt{3}+1}=\dfrac{{{\left( \sqrt{3} \right)}^{2}}+{{1}^{2}}+2\left( 1 \right)\left( \sqrt{3} \right)}{{{\left( \sqrt{3} \right)}^{2}}-{{1}^{2}}}$.
Simplifying the above equation, we have $\tan \left( {{75}^{\circ }} \right)=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}=\dfrac{{{\left( \sqrt{3} \right)}^{2}}+{{1}^{2}}+2\left( 1 \right)\left( \sqrt{3} \right)}{{{\left( \sqrt{3} \right)}^{2}}-{{1}^{2}}}=\dfrac{3+1+2\sqrt{3}}{3-1}=\dfrac{4+2\sqrt{3}}{2}$.
Thus, we have $\tan \left( {{75}^{\circ }} \right)=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}=\dfrac{4+2\sqrt{3}}{2}=2+\sqrt{3}$.
So, we observe that the given statement is incorrect.
Hence, the answer is False, which is option (b).

Note: We can also solve this question by using the trigonometric identity $\tan 2x=\dfrac{2\tan x}{1-{{\tan }^{2}}x}$ and then simplifying the expression using $\tan {{150}^{\circ }}=\dfrac{-1}{\sqrt{3}}$. Solve the quadratic equation by completing the square method or calculating the discriminant method.