BITSAT 2010
- Exam
- BITSAT
- Year
- 2010
- Subject
- Mathematics
- Chapter
- Trigonometric Equations
BITSAT 2010 · Mathematics · Trigonometric Equations
( 20^ ^ + 40^ ^ \ circ + 3 20^ ^ 40^ ^ ) is equal to
(\tan 20^{{}^{\circ}}+\tan 40^{{}^{\$circ}}+\sqrt{3} \tan20^{{}^{\circ}} \tan 40^{{}^{\circ}}) is equal to
Options
- A(/ 2)
- B(/ 4)
- C\$\sqrt{3}
- D1
Correct answer
C. \$\sqrt{3}
Step-by-step solution
$\begin{aligned}\sqrt{3}& =\tan 60^{{}^{\circ}}=\tan \left(40^{{}^{\circ}}+20^{{}^{\circ}}\right) \\
& =\frac{\tan 40^{{}^{\circ}}+\tan 20^{{}^{\circ}}}{1-\tan 40^{{}^{\circ}} \tan 20^{{}^{\circ}}}\\
\therefore \quad & \sqrt{3}-\sqrt{3} \tan40^{{}^{\circ}} \tan 20^{{}^{\circ}} \\
& =\tan 40^{{}^{\circ}}+\tan 20^{{}^{\circ}}
\end{aligned}$)
Hence (\tan 40^{{}^{\circ}}+\tan 20^{{}^{\circ}}+\sqrt{3} \tan40^{{}^{\circ}} \tan 20^{{}^{\circ}}) (=\sqrt{3})