Commerce (English Medium) इयत्ता ११ - CBSE Question Bank Solutions

 If \[\tan A = \frac{5}{6}\text{ and }\tan B = \frac{1}{11}\], prove that \[A + B = \frac{\pi}{4}\].

If \[\tan A = \frac{m}{m - 1}\text{ and }\tan B = \frac{1}{2m - 1}\], then prove that \[A - B = \frac{\pi}{4}\].

Find :

the 10^th term of the G.P.

\[\sqrt{2}, \frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}}, . . .\]

Prove that:
\[\cos^2 45^\circ - \sin^2 15^\circ = \frac{\sqrt{3}}{4}\]

Find the 4th term from the end of the G.P.

Prove that:
sin² (n + 1) A − sin² nA = sin (2n + 1) A sin A.

Prove that: \[\frac{\sin \left( A + B \right) + \sin \left( A - B \right)}{\cos \left( A + B \right) + \cos \left( A - B \right)} = \tan A\]

Prove that:
\[\frac{\sin \left( A - B \right)}{\cos A \cos B} + \frac{\sin \left( B - C \right)}{\cos B \cos C} + \frac{\sin \left( C - A \right)}{\cos C \cos A} = 0\]

Prove that:

Prove that:
sin² B = sin² A + sin² (A − B) − 2 sin A cos B sin (A − B)

Which term of the G.P. :

\[\sqrt{2}, \frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}}, \frac{1}{4\sqrt{2}}, . . . \text { is }\frac{1}{512\sqrt{2}}?\]

Prove that:
cos² A + cos² B − 2 cos A cos B cos (A + B) = sin² (A + B)

Which term of the G.P. :

\[2, 2\sqrt{2}, 4, . . .\text {  is }128 ?\]

Which term of the G.P.: `sqrt3, 3, 3sqrt3`, ... is 729?

Which term of the G.P. :

\[\frac{1}{3}, \frac{1}{9}, \frac{1}{27} . . \text { . is } \frac{1}{19683} ?\]

Prove that:
\[\frac{\tan \left( A + B \right)}{\cot \left( A - B \right)} = \frac{\tan^2 A - \tan^2 B}{1 - \tan^2 A \tan^2 B}\]

Which term of the progression 18, −12, 8, ... is \[\frac{512}{729}\] ?

Prove that:
tan 8x − tan 6x − tan 2x = tan 8x tan 6x tan 2x

Prove that:
\[\tan\frac{\pi}{12} + \tan\frac{\pi}{6} + \tan\frac{\pi}{12}\tan\frac{\pi}{6} = 1\]