Question

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

Answer 1

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.

Complete step-by-step answer:
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, ${\cos }^2{101}^0=1-{\sin }^2{101}^0$
Substitute these values in equation (i)
$\sqrt{{\cos }^2{101}^0}.\sec {101}^0=\cos {101}^0\sec {101}^0$
We know $\sec \theta ={\displaystyle \frac{1}{\cos \theta }}$
$\begin{array}{rl}& \therefore \sec {101}^0={\displaystyle \frac{1}{\cos {101}^0}}\\ & \therefore \cos {101}^0.\sec {101}^0={\cos }^0\times {\displaystyle \frac{1}{\cos {101}^0}}=1\end{array}$
Therefore, the correct answer is option C.

Note: Remember to use basic trigonometric formulas for the solution. You should learn or should know the formula and the inverse of trigonometric functions as the identities are important for all questions related to these formulas.