Question

# Solve $\sqrt{1-{{\sin }^{2}}{{101}^{0}}}.\sec {{101}^{0}}=$ A. 0B. -2 C. -1 D. 2

Hint: Put $1-{{\sin }^{2}}{{101}^{0}}={{\cos }^{2}}{{101}^{0}}$. Solve the rest using the trigonometric formula and identities. Substitute these formulas in the expression and simplify them until you get a whole number.

From the question $\sqrt{1-{{\sin }^{2}}{{101}^{0}}}.\sec {{101}^{0}}..................\left( i \right)$
We know ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$
$\Rightarrow {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta$
i.e, ${{\cos }^{2}}{{101}^{0}}=1-{{\sin }^{2}}{{101}^{0}}$
$\sqrt{{{\cos }^{2}}{{101}^{0}}}.\sec {{101}^{0}}=\cos {{101}^{0}}\sec {{101}^{0}}$
We know $\sec \theta =\dfrac{1}{\cos \theta }$
\begin{align} & \therefore \sec {{101}^{0}}=\dfrac{1}{\cos {{101}^{0}}} \\ & \therefore \cos {{101}^{0}}.\sec {{101}^{0}}={{\cos }^{0}}\times \dfrac{1}{\cos {{101}^{0}}}=1 \\ \end{align}