Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Given that tan (A+B)= tanA + tanB/ 1-tanA.tanB , find the value of tan 75 degree and tan 90 degree by taking suitable values of A and B.

Given that tan (A+B)= tanA + tanB/ 1-tanA.tanB , find the value of tan 75 degree and tan 90 degree by taking suitable values of A and B.

Answers (1)

\tan (A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \\\\ \text {Assume A = 45 degree, B = 30 degree }\\\\ \tan (45^{\circ} + 30^{\circ}) = \frac{\tan 45^{\circ} + \tan 30^{\circ}}{1 - \tan 45^{\circ} \tan 30^{\circ}} \\ \\ \tan (75^{\circ} ) = \frac{1+ \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} \\ \\ \tan (75^{\circ} ) = \frac{\sqrt{3} + 1}{\sqrt{3}-1} \\ \\ \tan (75^{\circ} ) = \frac{\sqrt{3} + 1}{\sqrt{3}-1} \times \frac{\sqrt{3} + 1}{\sqrt{3}+1} \\ \\ \tan (75^{\circ} ) = \frac{3+ 2\sqrt{3} + 1}{3-1} \\ \\ \tan (75^{\circ} ) = 2+ \sqrt{3} \\ \\ \text {Now assume A = 45 degree, B = 45 degree }\\\\ \tan (45^{\circ} + 45^{\circ}) = \frac{\tan 45^{\circ} + \tan 45^{\circ}}{1 - \tan 45^{\circ} \tan 45^{\circ}} \\ \\ \tan (90^{\circ} ) = \frac{1+ 1}{1 - 1 \times 1} \\ \\ \tan (90^{\circ} ) = \frac{2}{0} = \infty \\ \\

Posted by

Ravindra Pindel

View full answer

Similar Questions

Latest Question