MCQ
$\sec^4\text{A}-\sec^2\text{A}$ is equal to:
- A$\tan^2\text{A}-\tan^4\text{A}$
- B$\tan^4\text{A}-\tan^2\text{A}$
- ✓$\tan^4\text{A}+\tan^2\text{A}$
- D$\tan^2\text{A}+\tan^4\text{A}$
✓
Answer
Correct option: C.
$\tan^4\text{A}+\tan^2\text{A}$
$\sec^4-\sec^2\text{A}=\sec^2\text{A}(\sec^2\text{A}-1)$
$=(1+\tan^2\text{A})\tan^2\text{A}$
$\begin{cases}\sec^2\text{A}=1+\tan^2\text{A}\\
\sec^2\text{A}-1=\tan^2\text{A}\end{cases}$
$=\tan^2\text{A}+\tan^4\text{A}$
$=\tan^4\text{A}+\tan^2\text{A}$
$=(1+\tan^2\text{A})\tan^2\text{A}$
$\begin{cases}\sec^2\text{A}=1+\tan^2\text{A}\\
\sec^2\text{A}-1=\tan^2\text{A}\end{cases}$
$=\tan^2\text{A}+\tan^4\text{A}$
$=\tan^4\text{A}+\tan^2\text{A}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
If $7\tan\theta=4$ then $\frac{(7\sin\theta-3\cos\theta)}{(7\sin\theta+3\cos\theta)} =?$
→The pair of equations $x + 2y + 5 = 0$ and $-3x - 6y + 1 = 0$ has:
→The points $P(0, 6), Q(-5, 3)$ and $R(3, 1)$ are the vertices of a triangle, which is:
→If $\triangle\text{ABC}$ is an equilateral triangle such that $\text{AD}\perp\text{BC},$ then $AD^2 =$
→If the equation $x^2 - bx + 1 = 0$ does not possess real roots, then:
→Which of the following is not a measure of central tendency?
In the given figure, three circles with centres $\text{A, B, C,}$ respectively, touch each other externally. If $\text{AB} = 5\ cm, \text{BC} = 7\ cm$ and $\text{CA} = 6\ cm,$ the radius of the circle with centre $A$ is:
→Which of the following cannot be the probability of an event?
If $\cot \theta=\frac{3}{4}$, then $\tan \theta=$ ______
→$\tan5^\circ\tan25^\circ\tan30^\circ\tan65^\circ\tan85^\circ=?$
→