Question

Find the value of \[{\tan ^2}{60^0} + 2{\tan ^2}{45^0}\].

Answer 1

Hint: First of all, write the given equation and substitute the values of tangent angles as we have learnt from the trigonometric ratio table to obtain the required answer. So, use this concept to reach the solution of the given problem.

Completes step-by-step solution -
Given equation is \[{\tan ^2}{60^0} + 2{\tan ^2}{45^0}\]
We know that \[\tan {60^0} = \sqrt 3 \] and \[\tan {45^0} = 1\]
By substituting these values, we get
\[
  {\tan ^2}{60^0} + 2{\tan ^2}{45^0} = {\left( {\tan {{60}^0}} \right)^2} + 2{\left( {\tan {{45}^0}} \right)^2} \\
  {\tan ^2}{60^0} + 2{\tan ^2}{45^0} = {\left( {\sqrt 3 } \right)^2} + 2{\left( 1 \right)^2} \\
  {\tan ^2}{60^0} + 2{\tan ^2}{45^0} = \sqrt 3 \times \sqrt 3 + 2 \times 1 \\
  {\tan ^2}{60^0} + 2{\tan ^2}{45^0} = 3 + 2 \\
  \therefore {\tan ^2}{60^0} + 2{\tan ^2}{45^0} = 5 \\
\]
Thus, the value of \[{\tan ^2}{60^0} + 2{\tan ^2}{45^0}\] is 5.
Note: The trigonometric table ratio is made up of trigonometric ratios that are interrelated to each other-sine, cosine, tangent, cosecant, secant, cotangent. Here we have used the values of \[\tan {60^0} = \sqrt 3 \], \[\tan {45^0} = 1\].