Question
If $\tan\theta=\frac{1}{\sqrt{7}},$ show that $\frac{\big(\text{coses}^2\theta-\sec^2\theta\big)}{\big(\text{cosec}^2\theta+\sec^2\theta\big)}=\frac34.$
✓
Answer
Given: $\tan\theta=\frac{\text{BC}}{\text{AB}}=\frac{1}{\sqrt{7}}$
Let BC = 1k and $\text{AB}=\sqrt{7}\text{k}$
Where k is positive
Let us draw a $\triangle\text{ABC}$ in which $\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$
By pythagoras theorem, we have
$\text{AC}^2 = (\text{AB}^2 + \text{BC}^2)$
$\Rightarrow\text{AC}^2=\Big[\big(\sqrt{7}\text{k}\big)^2+\big(1\text{k}\big)^2\Big]$
$= 7\text{k}^2 + 1\text{k}^2 = 8\text{k}^2$
$\Rightarrow\text{AC}=2\sqrt{2}\text{k}$
$\text{cosec}\theta=\frac{\text{AC}}{\text{BC}}=\frac{2\sqrt{2}\text{k}}{1\text{k}}={2\sqrt{2}}$
$\sec\theta=\frac{\text{AC}}{\text{AB}}=\frac{2\sqrt{2}\text{k}}{\sqrt{7}\text{k}}=\frac{2\sqrt{2}}{\sqrt{7}}$
$\Rightarrow\frac{\big(\text{cosec}^2\theta-\sec^2\theta\big)}{\big(\text{cosec}^2\theta+\sec^2\theta\big)}=\begin{bmatrix}\frac{\big(2\sqrt{2}\big)^2-\Big(\frac{2\sqrt{2}}{\sqrt{7}}\Big)^2}{\big(2\sqrt{2}\big)^2+\Big(\frac{2\sqrt{2}}{\sqrt{7}}\Big)^2}\end{bmatrix}$
$=\frac{\big(8-\frac{8}{7}\big)}{\big(8+\frac87\big)}=\frac{\big(\frac{48}{7}\big)}{\big(\frac{64}{7}\big)}=\frac{48}{64}=\frac34$
Hence, $\Big(\frac{\text{cosec}^2\theta-\sec^2\theta}{\text{cosec}^2\theta+\sec^2\theta}\Big)=\frac34$
Let BC = 1k and $\text{AB}=\sqrt{7}\text{k}$
Where k is positive
Let us draw a $\triangle\text{ABC}$ in which $\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$
By pythagoras theorem, we have
$\text{AC}^2 = (\text{AB}^2 + \text{BC}^2)$
$\Rightarrow\text{AC}^2=\Big[\big(\sqrt{7}\text{k}\big)^2+\big(1\text{k}\big)^2\Big]$
$= 7\text{k}^2 + 1\text{k}^2 = 8\text{k}^2$
$\Rightarrow\text{AC}=2\sqrt{2}\text{k}$
$\text{cosec}\theta=\frac{\text{AC}}{\text{BC}}=\frac{2\sqrt{2}\text{k}}{1\text{k}}={2\sqrt{2}}$
$\sec\theta=\frac{\text{AC}}{\text{AB}}=\frac{2\sqrt{2}\text{k}}{\sqrt{7}\text{k}}=\frac{2\sqrt{2}}{\sqrt{7}}$
$\Rightarrow\frac{\big(\text{cosec}^2\theta-\sec^2\theta\big)}{\big(\text{cosec}^2\theta+\sec^2\theta\big)}=\begin{bmatrix}\frac{\big(2\sqrt{2}\big)^2-\Big(\frac{2\sqrt{2}}{\sqrt{7}}\Big)^2}{\big(2\sqrt{2}\big)^2+\Big(\frac{2\sqrt{2}}{\sqrt{7}}\Big)^2}\end{bmatrix}$
$=\frac{\big(8-\frac{8}{7}\big)}{\big(8+\frac87\big)}=\frac{\big(\frac{48}{7}\big)}{\big(\frac{64}{7}\big)}=\frac{48}{64}=\frac34$
Hence, $\Big(\frac{\text{cosec}^2\theta-\sec^2\theta}{\text{cosec}^2\theta+\sec^2\theta}\Big)=\frac34$
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
Find the roots of the following equations, if they exist, by applying the quadratic formula:$2\text{x}^2+5\sqrt3\text{x}+6=0$
→Solve the following system of equations by the method of cross-multiplication:
→$2ax + 3by = a + 2b,$
$3ax + 2by = 2a + b.$
In the given fingure, $\triangle\text{ABC}$ is right-angled at A. semicircles are drawn on AB, AC and BC as diameter, it is given that AB = 3cm and AC = 4cm. find the area of the shaded region.
→A bag contains 24 balls of which x are red, 2x are white and 3x are blue. A ball is selected at random. What is the probability that it is:
- not red?
- white?
Solve the given pair of linear equation by the elimination method and the substitution method: 3x + 4y = 10 and 2x – 2y = 2
How many spherical lead shots each of diameter 4.2cm can be obtained from a solid rectangular lead piece with dimensions 66cm, 42cm and 21cm.
Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients:
$4x^2 - 4x + 1$
→$4x^2 - 4x + 1$
In Fig., $\text{DEFG}$ is a square in a triangle $\text{ABC}$ right angled at $A$ . Prove that
$i. \triangle AGF \sim \triangle DBG$
$ii. \triangle AGF \sim \triangle EFC$
→$i. \triangle AGF \sim \triangle DBG$
$ii. \triangle AGF \sim \triangle EFC$
Find the value of a, so that the point (3, a) lies on the line respresented by 2x - 3y = 5.
If AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as shown in Figure. Prove that $\angle\text{BAT}=\angle\text{ACB}.$
→