MCQ
If $\tan\theta=\frac{1}{\sqrt{7}},$ then $\frac{\text{cosec}^2\theta-\sec^2\theta}{\text{cosec}^2\theta+\sec^2\theta}=$
- A$\frac{5}{7}$
- B$\frac{3}{7}$
- C$\frac{1}{12}$
- ✓$\frac{3}{4}$
✓
Answer
Correct option: D.
$\frac{3}{4}$
Given that :
$\tan\theta=\frac{1}{\sqrt{7}}$
We are asked to find the value of the following expression
$\frac{\text{cosec}^2\theta-\sec^2\theta}{\text{cosec}^2\theta+\sec^2\theta}$
Since $\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}$
$\Rightarrow{\text{perpendicular}}=1$
$\Rightarrow{\text{Base}}=\sqrt{7}$
$\Rightarrow{\text{Hypotennuse}}=\sqrt{1+7}$
$\Rightarrow{\text{Hypotennuse}}=\sqrt{8}$
We know that $\sec\theta=\frac{\text{Hypotenuse}}{\text{Base}}$ and $\text{cosec}\theta=\frac{\text{Hypotenuse}}{\text{perpendicular}}$
We find :
$\frac{\text{cosec}^2\theta-\sec^2\theta}{\text{cosec}^2\theta+\sec^2\theta}$
$=\frac{\Big(\frac{\sqrt{8}}{1}\Big)^2-\Big(\frac{\sqrt{8}}{\sqrt{7}}\Big)^2}{\Big(\frac{\sqrt{8}}{1}\Big)^2+\Big(\frac{\sqrt{8}}{\sqrt{7}}\Big)^2}$
$=\frac{\frac{{8}}{1}-\frac{{8}}{{7}}}{\frac{{8}}{1}+\frac{{8}}{{7}}}$
$=\frac{\frac{48}{7}}{\frac{64}{7}}$
$=\frac{3}{4}$
Hence the correct option is $(d)$
$\tan\theta=\frac{1}{\sqrt{7}}$
We are asked to find the value of the following expression
$\frac{\text{cosec}^2\theta-\sec^2\theta}{\text{cosec}^2\theta+\sec^2\theta}$
Since $\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}$
$\Rightarrow{\text{perpendicular}}=1$
$\Rightarrow{\text{Base}}=\sqrt{7}$
$\Rightarrow{\text{Hypotennuse}}=\sqrt{1+7}$
$\Rightarrow{\text{Hypotennuse}}=\sqrt{8}$
We know that $\sec\theta=\frac{\text{Hypotenuse}}{\text{Base}}$ and $\text{cosec}\theta=\frac{\text{Hypotenuse}}{\text{perpendicular}}$
We find :
$\frac{\text{cosec}^2\theta-\sec^2\theta}{\text{cosec}^2\theta+\sec^2\theta}$
$=\frac{\Big(\frac{\sqrt{8}}{1}\Big)^2-\Big(\frac{\sqrt{8}}{\sqrt{7}}\Big)^2}{\Big(\frac{\sqrt{8}}{1}\Big)^2+\Big(\frac{\sqrt{8}}{\sqrt{7}}\Big)^2}$
$=\frac{\frac{{8}}{1}-\frac{{8}}{{7}}}{\frac{{8}}{1}+\frac{{8}}{{7}}}$
$=\frac{\frac{48}{7}}{\frac{64}{7}}$
$=\frac{3}{4}$
Hence the correct option is $(d)$
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
Identify the center of the circle :
$\triangle\text{ABC}\sim\triangle\text{DEF}.$ If $BC = 3\ cm, EF = 4\ cm$ and $\text{ar}(\triangle\text{ABC})=54\text{ cm}^2,$ then $\text{ar}(\triangle\text{DEF}):$
→Choose the correct answer from the given four options : If in two traingles $\text{ABC}$ and $\text{PQR},$
$\frac{\text{AB}}{\text{QR}}=\frac{\text{BC}}{\text{PR}}=\frac{\text{CA}}{\text{PQ}},$ then :
→$\frac{\text{AB}}{\text{QR}}=\frac{\text{BC}}{\text{PR}}=\frac{\text{CA}}{\text{PQ}},$ then :
The mean of $2,7,6$ and $x$ is 15 and the mean of $18,1,6, x$ and $y$ is 10 . What is the value of $y$ ?
→A die is thrown once. Find the probability of getting a number less than 7.
The product of two consecutive even integers is 528. The quadratic equation corresponding to the above statement, is
While computing mean of grouped data, we assume that the frequencies are :
The maximum value of $\frac{1}{\operatorname{cosec} \theta}$ is
→A letter of English alphabets is chosen at random. The probability that the letter chosen is a vowel is :
A polynomial of degree $..........$ is called a quadratic polynomial :
→