Question

The value of $\csc 15{}^\circ +\sec 15{}^\circ =$
A. $2\sqrt{3}$
B. $\sqrt{6}$
C. $2\sqrt{6}$
D. $\sqrt{6}+\sqrt{2}$

Answer 1

The angle $15{}^\circ$ can be written as $15{}^\circ =45{}^\circ -30{}^\circ$ .
We need to know that $\csc A=\dfrac{1}{\sin A}$ and $\sec A=\frac{1}{\cos A}$ .
Some useful Trigonometric Identities:
$\sin (A\pm B)=\sin A\cos B\pm \sin B\cos A$
$\cos (A\pm B)=\cos A\cos B\mp \sin A\sin B$

Values of Trigonometric Ratios for Common Angles:

| $0{}^\circ$| $30{}^\circ$| $45{}^\circ$ | $60{}^\circ$| $90{}^\circ$
sin| 0| $\dfrac{1}{2}$| $\dfrac{1}{\sqrt2}$ | $\dfrac{\sqrt3}{2}$ | 1
cos| 1| $\dfrac{\sqrt3}{2}$| $\dfrac{1}{\sqrt2}$ | $\dfrac{1}{2}$ | 0
tan| 0| $\dfrac{1}{\sqrt3}$| 1| $\sqrt3$ | $\infty$
csc| $\infty$| 2| $\sqrt2$ | $\dfrac{2}{\sqrt3}$ | 1
sec| 1| $\dfrac{2}{\sqrt3}$ | $\sqrt2$ | 2| $\infty$
cot| $\infty$ | $\sqrt3$ | 1| $\dfrac{1}{\sqrt3}$ | 0

Complete step-by-step answer:
Let us first calculate the values of $\sin 15{}^\circ$ and $\cos 15{}^\circ$ .
Using the identity $\sin (A-B)=\sin A\cos B-\sin B\cos A$ , we get:
$\sin 15{}^\circ =\sin \left( 45{}^\circ -30{}^\circ \right)=\sin 45{}^\circ \cos 30{}^\circ -\sin 30{}^\circ \cos 45{}^\circ$
Substituting the values of $\sin 45{}^\circ$ , $\cos 30{}^\circ$ , etc.
⇒ $\sin 15{}^\circ =\left( \dfrac{1}{\sqrt{2}} \right)\left( \dfrac{\sqrt{3}}{2} \right)-\left( \dfrac{1}{2} \right)\left( \dfrac{1}{\sqrt{2}} \right)$
⇒ $\sin 15{}^\circ =\dfrac{\sqrt{3}-1}{2\sqrt{2}}$
And, using the identity $\cos (A-B)=\cos A\cos B+\sin A\sin B$ , we get:
$\cos 15{}^\circ =\cos \left( 45{}^\circ -30{}^\circ \right)=\cos 45{}^\circ \cos 30{}^\circ +\sin 45{}^\circ \sin 30{}^\circ$
Substituting the values of $\sin 45{}^\circ$ , $\cos 30{}^\circ$ , etc.
⇒ $\cos 15{}^\circ =\left( \dfrac{1}{\sqrt{2}} \right)\left( \dfrac{\sqrt{3}}{2} \right)+\left( \dfrac{1}{\sqrt{2}} \right)\left( \dfrac{1}{2} \right)$
⇒ $\cos 15{}^\circ =\dfrac{\sqrt{3}+1}{2\sqrt{2}}$
Now, $\csc 15{}^\circ +\sec 15{}^\circ =\dfrac{1}{\sin 15{}^\circ }+\dfrac{1}{\cos 15{}^\circ }$ . Substituting the values of $\sin 15{}^\circ$ and $\cos 15{}^\circ$ from above, we get:
$\csc 15{}^\circ +\sec 15{}^\circ =\dfrac{1}{\left( \dfrac{\sqrt{3}-1}{2\sqrt{2}} \right)}+\dfrac{1}{\left( \dfrac{\sqrt{3}+1}{2\sqrt{2}} \right)}$
= $\dfrac{2\sqrt{2}}{\sqrt{3}-1}+\dfrac{2\sqrt{2}}{\sqrt{3}+1}$
On equating the denominators, we get:
= $\dfrac{2\sqrt{2}\left( \sqrt{3}+1 \right)}{\left( \sqrt{3}-1 \right)\left( \sqrt{3}+1 \right)}+\dfrac{2\sqrt{2}\left( \sqrt{3}-1 \right)}{\left( \sqrt{3}+1 \right)\left( \sqrt{3}-1 \right)}$
On multiplying the terms in the numerator and denominators, we get:
= $\dfrac{2\sqrt{6}+\sqrt{2}}{3+\sqrt{3}-\sqrt{3}-1}+\dfrac{2\sqrt{6}-\sqrt{2}}{3-\sqrt{3}+\sqrt{3}-1}$
= $\dfrac{2\sqrt{6}+\sqrt{2}}{2}+\dfrac{2\sqrt{6}-\sqrt{2}}{2}$
= $\dfrac{4\sqrt{6}}{2}$
= $2\sqrt{6}$ .
Hence, the correct answer option is C. $2\sqrt{6}$ .

Note: (i) The values of trigonometric ratios for the common angles $0{}^\circ ,30{}^\circ ,45{}^\circ ,60{}^\circ ,90{}^\circ$ are calculated by using Pythagoras's theorem and properties of isosceles triangles for $60{}^\circ -60{}^\circ -60{}^\circ$ and $90{}^\circ -45{}^\circ -45{}^\circ$ triangles.
(ii)If we know the values of trigonometric ratios for the common angles $0{}^\circ ,30{}^\circ ,45{}^\circ ,60{}^\circ ,90{}^\circ$ , we can calculate the values of trigonometric ratios for angles which are sum or differences of these, like $15{}^\circ ,75{}^\circ ,105{}^\circ ,120{}^\circ ,135{}^\circ$ .
(iii)We see that $\sin 2A=2\sin A\cos A$ . Now, we can also find the values of $\sin 3A$ by using the fact that $3A = 2A + A$ , and then $\sin 5A$ as $5A = 3A + 2A$ , and so on.
(iv)Since, ${{\sin }^{2}}A+{{\cos }^{2}}A=1$ , we can always calculate $\cos A$ from $\sin A$ , and then the other ratios follow.