^260^ -1=? - Vidyadip

MCQ

$\sec^260^\circ-1=?$

A
$2$
✓
$3$
C
$4$
D
$0$

✓

Answer

Correct option: B.

$3$

$\sec^260^\circ-1=(2)^2-1$
$=4-1$
$=3$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Trigonometric Identities→Trigonometric Identities — all question types→Maths STD 10 question papers→Maharashtra Board STD 10 question papers→Maharashtra Board question paper generator→

Similar questions

If $\theta$ is an acute angle such that $\sec^2\theta=3,$ then the value of $\frac{\tan^2\theta-\text{cosec}^2\theta}{\tan^2\theta+\text{cosec}^2\theta}$ is:

In an A.P. first two terms are – 3, 4 then 21^st term is . . .

If $\sin \theta+\cos \theta= a$, and $\sin \theta-\cos \theta= b$, then $=$ ______

For some integer $m,$ every even integer is of the form:

The $\text{HCF}$ of two numbers is $27$ and their $\text{LCM}$ is $162.$ If one of the numbers is $54,$ what is the other number?

$(\text{cosec }\theta-\sin\theta)(\sec\theta-\cos\theta)(\tan\theta+\cot\theta)$ is equal to:

If $\text{a}\cos\theta+\text{b}\sin\theta=\text{m}$ and $\text{a}\sin\theta-\text{b}\cos\theta=\text{n},$ then $a^2 + b^2 =$

The coordinates of the point $P$ dividing the line segment joining the points $A(1, 3)$ and $B(4, 6)$ in the ratio $2 : 1$ are:

In ∆ABC, AB =$6 \sqrt{3}$ cm, AC = 12 cm, and BC = 6 cm then $m \angle A$ = ?

If $A$ and $B$ are acute angles such that $\sin\text{A}=\cos\text{B}$ then $(\text{A}+\text{B})=?$