NIMCET 2007 — Mathematics PYQ

To find the exact value of $\sec \frac{7\pi }{12}$, we can first evaluate $\cos \frac{7\pi }{12}$ and then take its reciprocal.

Step 1: Simplify the angle

The given angle is $\frac{7\pi }{12}$. We can split this angle into a sum of two standard angles whose trigonometric values are known:

$$\frac{7\pi }{12}=\frac{3\pi }{12}+\frac{4\pi }{12}=\frac{\pi }{4}+\frac{\pi }{3}$$

Step 2: Find the value of $\cos \frac{7\pi }{12}$

Using the cosine compound angle formula $\cos (x+y)=\cos x\cos y-\sin x\sin y$:

$$\cos \left(\frac{\pi }{4}+\frac{\pi }{3}\right)=\cos \frac{\pi }{4}\cos \frac{\pi }{3}-\sin \frac{\pi }{4}\sin \frac{\pi }{3}$$

Substitute the standard values ($\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$, $\cos \frac{\pi }{3}=\frac{1}{2}$, $\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}$, $\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$):

$$\cos \frac{7\pi }{12}=\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)-\left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right)$$

$$\cos \frac{7\pi }{12}=\frac{1}{2\sqrt{2}}-\frac{\sqrt{3}}{2\sqrt{2}}=\frac{1-\sqrt{3}}{2\sqrt{2}}$$

Step 3: Calculate $\sec \frac{7\pi }{12}$ by taking the reciprocal

Since $\sec \theta =\frac{1}{\cos \theta }$:

$$\sec \frac{7\pi }{12}=\frac{2\sqrt{2}}{1-\sqrt{3}}$$

Step 4: Rationalize the denominator

To match the options provided in image_21dde5.png, multiply the numerator and the denominator by the conjugate $(1+\sqrt{3})$:

$$\sec \frac{7\pi }{12}=\frac{2\sqrt{2}(1+\sqrt{3})}{(1-\sqrt{3})(1+\sqrt{3})}$$

Using the algebraic identity $(a-b)(a+b)=a^2-b^2$ for the denominator:

$$\sec \frac{7\pi }{12}=\frac{2\sqrt{2}(\sqrt{3}+1)}{1-3}$$

$$\sec \frac{7\pi }{12}=\frac{2\sqrt{2}(\sqrt{3}+1)}{-2}$$

Canceling out the common factor $2$:

$$\sec \frac{7\pi }{12}=-\sqrt{2}(\sqrt{3}+1)$$

Conclusion

The exact value matches option (c).

Correct Option: (c) $-\sqrt{2}(\sqrt{3}+1)$