Question

If $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$, then the minimum value of ${\cos }^3\theta +{\sec }^3\theta$ is

• A. $1$
• B. $2$
• C. $3$
• D. None of the above.

Answer 1

Answer: (B)

$2$

To find the optimum, we take the derivative and equate to zero.

given, $co{s}^3\theta +se{c}^3\theta$

taking derivative w.r.to $\theta$ and equating to zero we get,

$⟹3co{s}^2\theta (-sin\theta )+3se{c}^2\theta (sec\theta \times tan\theta )=0$

$⟹co{s}^2\theta (-sin\theta )+se{c}^2\theta (sec\theta \times \frac{sin\theta }{cos\theta })=0$

$⟹co{s}^2\theta (-sin\theta )+se{c}^2\theta (sec\theta \times sin\theta \times sec\theta )=0$

$⟹co{s}^2\theta (-sin\theta )+se{c}^4\theta \left(sin\theta \right)=0$

$⟹sin\theta (-co{s}^2\theta +se{c}^4\theta )=0$

$⟹sin\theta =0$ and $(-co{s}^2\theta +se{c}^4\theta )=0$

since, $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$

Therefore, $sin\theta =0⟹\theta =0^o$

$-co{s}^2\theta +se{c}^4\theta =0$

$⟹-co{s}^2\theta +\frac{1}{co{s}^4\theta }=0$

$⟹\frac{1-co{s}^6\theta }{co{s}^4\theta }=0$

$⟹1-co{s}^6\theta =0$

$⟹co{s}^6\theta =1⟹cos\theta =1$

since, $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$

Therefore, $cos\theta =1⟹\theta =0^o$

Substituting $\theta =0^o$ we get,

$(cos0{)}^3+(sec0{)}^3=1+1=2$