Question
Prove the following identites:
$\big(\sec^2\theta-1\big)\big(\text{cosec}^2\theta-1\big)=1$
$\big(\sec^2\theta-1\big)\big(\text{cosec}^2\theta-1\big)=1$
✓
Answer
$\text{L.H.S.}=\big(\sec^2\theta-1\big)\big(\text{cosec}^2\theta-1\big)$
$=\tan^2\theta\times\cot^2\theta$ $\begin{bmatrix}\because\big(\sec^2\theta-1\big)=\tan^2\theta,\\\big(\text{cosec}^2\theta-1\big)=\cot^2\theta\end{bmatrix}$
$=\tan^2\theta\times\frac{1}{\tan^2\theta}$
$=1$
$=\text{R.H.S.}$
$\therefore\text{LHS}=\text{RHS}$
$=\tan^2\theta\times\cot^2\theta$ $\begin{bmatrix}\because\big(\sec^2\theta-1\big)=\tan^2\theta,\\\big(\text{cosec}^2\theta-1\big)=\cot^2\theta\end{bmatrix}$
$=\tan^2\theta\times\frac{1}{\tan^2\theta}$
$=1$
$=\text{R.H.S.}$
$\therefore\text{LHS}=\text{RHS}$
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
Five cards the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
- What is the probability that the card is the queen?
- If the queen is drawn and put a side, what is the probability that the second card picked up is:
- An ace?
- A queen?
If the centroid of the triangle formed by points $P ( a , b ), Q ( b , c )$ and $R ( c , a )$ is at the origin, what is the value of $a + b +$ c?
→Very-Short and Short-Answer Questions:
If $\frac{\text{x}}{4}+\frac{\text{y}}{\text{3}}=\frac{5}{\text{12}}$ and $\frac{\text{x}}{\text{2}}+\text{y}=1$ then find the value of (x + y).
→If $\frac{\text{x}}{4}+\frac{\text{y}}{\text{3}}=\frac{5}{\text{12}}$ and $\frac{\text{x}}{\text{2}}+\text{y}=1$ then find the value of (x + y).
Determine the nature of roots for each of the quadratic equation.
3x2 – 5x + 7 = 0
3x2 – 5x + 7 = 0
Two tangents TP and TQ are drawn from an external point T to a circle with centre O . If they are inclined to each other at an angle of 100°, then what is the value of $\angle\text{POQ}$?
→From adjoining figure, $\angle ABC =90^{\circ}, \angle DCB =90^{\circ}, AB =6, DC =8$, then $\frac{ A (\triangle ABC )}{ A (\triangle BCD )}=$ ?
→See fig 2.11. In $\triangle \mathrm{ABC}, \angle \mathrm{B}=90^{\circ}, \angle \mathrm{A}=30^{\circ}, \mathrm{AC}=14$, then find $\mathrm{AB}$ and $\mathrm{BC}$
→
Find the mode of the following distribution.
|
Class-interval
|
25-30
|
30-35
|
35-40
|
40-45
|
45-50
|
50-60
|
|
Frequency
|
25
|
34
|
50
|
42
|
38
|
14
|
A bag contains 50 cards. Each card bears only one number from 1 to 50. One
card is drawn at randomfrom the bag. Write the sample space. Also write the
events A, B and find the number of sample points in them.
(i) Condition for event A : the number on the card is divisible by 6.
(ii) Condition for event B : the number on the card is a complete square.
card is drawn at randomfrom the bag. Write the sample space. Also write the
events A, B and find the number of sample points in them.
(i) Condition for event A : the number on the card is divisible by 6.
(ii) Condition for event B : the number on the card is a complete square.
$ABC$ is an isosceles triangle, right-angled at $B$. Similar triangle $ACD$ and $ABE$ are constructed on sides $AC$ and $AB$. Find the ratio between the areas of $\triangle\text{ABE}$ and $\triangle\text{ACD}.$
→