Question

Prove that: cos 11 + sin 11 % cos 11 - sin 11 = tan 56​

Answer 1

So we are given $\dfrac{\text{cos 11 + sin 11}}{\text{cos 11 - sin 11}}$ in the L.H.S. and tan 56 in the R.H.S.

So, let's start solving the question from the L.H.S.

$\dfrac{\text{cos 11 + sin 11}}{\text{cos 11 - sin 11}}$

Now, let's start by dividing the numerator and the denominator by cos 11.

$\dfrac{\dfrac{\text{cos 11 + sin 11}}{\text{cos 11}}}{\dfrac{\text{cos 11 - sin 11}}{\text{cos 11}}}$

Simplify.

$\dfrac{\dfrac{\text{cos 11}}{\text{cos 11}}+\dfrac{\text{sin 11}}{\text{cos 11}}}{\dfrac{\text{cos 11}}{\text{cos 11}}-\dfrac{\text{sin 11}}{\text{cos 11}}}\\\\\\\implies \dfrac{1+\dfrac{\text{sin 11}}{\text{cos 11}}}{1-\dfrac{\text{sin 11}}{\text{cos 11}}}\\\\\\\implies \dfrac{1+\text{tan 11}\textdegree}{1-\text{tan 11}\textdegree}$

Now as we know that "tan 45° = 1", we'll replace the 1 to tan 45°.

$\dfrac{(\text{tan 45}\textdegree)+\text{tan 11}\textdegree}{1-(\text{tan 45}\textdegree)(\text{tan 11}\textdegree)}$

Now let's simplify.

$\dfrac{\text{tan (a)}+\text{tan (b)}}{1-\text{tan (a)}\text{tan (b)}}\\\\\implies \text{tan (a + b)}$

Now let's put the value of a and b in the above equation;

tan (45° +11°) = tan (56°) = R.H.S.

                                 Hence, Proved