Download App

Trigonometric Functions — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions1 Mark
MCQ
If $\sec\text{x}=\text{x}+\frac{1}{4\text{x}},$ then $\sec\text{x}+\tan\text{x}=$
  • A
    $\text{x},\frac{1}{\text{x}}$
  • ✓
    $2\text{x},\frac{1}{2\text{x}}$
  • C
    $-2\text{x},\frac{1}{2\text{x}}$
  • D
    $-\frac{1}{\text{x}},\text{x}$
✓

Answer

Correct option: B.
$2\text{x},\frac{1}{2\text{x}}$
We have :
$\sec\text{x} = \text{x} +\frac{1}{4\text{x}}$
$\Rightarrow\sec^2\text{x}=\text{x}^2+\frac{1}{16\text{x}^2}+\frac{1}{2}$
$\Rightarrow1+\tan^2\text{x}$
$=1+\text{x}^2+\frac{1}{16\text{x}^2}-\frac{1}{2}$
$\Rightarrow\tan^2\text{x}=\text{x}^2+\frac{1}{16\text{x}^2}-\frac{1}{2}$
$\Rightarrow\tan^2\text{x}=\Big(\text{x}-\frac{1}{4\text{x}}\Big)^2$
$\therefore\tan\text{x}=\pm\Big(\text{x}-\frac{1}{4\text{x}}\Big)$
$\sec\text{x}-\tan\text{x}=\Big(\text{x}+\frac{1}{4\text{x}}\Big)-\Big(\text{x}-\frac{1}{4\text{x}}\Big)\text{or}$
$=\Big(\text{x}+\frac{1}{4\text{x}}\Big)-\Big[-\Big(\text{x}-\frac{1}{4\text{x}}\Big)\Big]$
$=\frac{1}{2\text{x}}$ or $2\text{x}$

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 "MCQ" from Trigonometric Functions→Trigonometric Functions — all question types→Maths STD 11 Science question papers→Maharashtra Board STD 11 Science question papers→Maharashtra Board question paper generator→

Similar questions

If $f (x)=x^2+\frac{1}{x}, x \neq 0$ then $f \left(\frac{1}{x}\right)=$
→
Value of $\left(\frac{-1+\sqrt{-3}}{2}\right)^{40}+\left(\frac{-1-\sqrt{-3}}{2}\right)^{40}$ is
→
If $\text{z}=\frac{1}{(1-\text{i})(2+3\text{i})},$ then $|\text{z}|=$
→
The equation of the directrix of the parabola $x^2+4 x+2 y=0$ is
→
If $P ( A )= P ( B )=x$ and $P ( A \cap B )= P \left( A ^{\prime} \cap B ^{\prime}\right)=\frac{1}{3}$, then $x=$
→
$\lim _{x \rightarrow 3}\left(\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\right)=$
→
Consider the circle $x^2+y^2-6 x+4 y=12$. The equation of a tangent to this circle that is parallel to the line $4 x+3 y+5=0$ is
→
Three integers are chosen at random from the first $20$ integers. The probability that their product is even is :
→
$\frac{(1+i)^{2016}}{(1-i)^{2014}}=$
→
The value of $\lim _{x \rightarrow \infty} \frac{(x+1)(3 x+4)}{x^2(x-8)}$ is equal to
→