Question
Very-short and Short-Answer Questions.
Write the value of $\text{cosec}^2\theta(1+\cos\theta)(1-\sin\theta).$
Write the value of $\text{cosec}^2\theta(1+\cos\theta)(1-\sin\theta).$
✓
Answer
$\text{cosec}^2\theta(1+\cos\theta)(1-\sin\theta)$$=\text{cosec}^2\theta\big(1-\cos^2\theta\big)$
$=\text{cosec}^2\theta\times\sin^2\theta$
$=\frac{1}{\sin^2\theta}\times\sin^2\theta$
$=1$
$=\text{cosec}^2\theta\times\sin^2\theta$
$=\frac{1}{\sin^2\theta}\times\sin^2\theta$
$=1$
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 $\sin^2\theta\cos^2\theta(1+\tan^2\theta)(1+\cos^2\theta)=\lambda,$ then find is the value of $\lambda$.
→Show that the progressions given below is an AP. Find the first term, common difference and next term.
$\sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32},\ ....$
→$\sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32},\ ....$
On comparing the ratios $\frac{\text{a}_1}{\text{a}_2},\frac{\text{b}_1}{\text{b}_2}$ and $\frac{\text{c}_1}{\text{c}_2},$ and without drawing them, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincide.
→$6x - 3y + 10 = 0$
$2x - y + 9 = 0$
Prove that $(\text{cosec A}-\cot\text{A})^2=\frac{(1-\cos\text{A})}{(1+\cos\text{A})}.$
→Very-Short and Short-Answer Qustions:
If $\cot\theta=\frac{1}{\sqrt{3}},$ write the value of $\frac{\big(1-\cos^2\theta\big)}{\big(2-\sin^2\theta\big)}.$
→If $\cot\theta=\frac{1}{\sqrt{3}},$ write the value of $\frac{\big(1-\cos^2\theta\big)}{\big(2-\sin^2\theta\big)}.$
Find the sum:
$18+15\frac{1}{2}+13\ + ..... \ +\Big(-49\frac{1}{2}\Big)$
→$18+15\frac{1}{2}+13\ + ..... \ +\Big(-49\frac{1}{2}\Big)$
In a $\triangle\text{ABC},$ AB = BC = CA = 2a and $\text{AD}\perp\text{BC}.$ Prove that
Area $(\triangle\text{ABC})=\sqrt{3}\text{ a}^2$
→Area $(\triangle\text{ABC})=\sqrt{3}\text{ a}^2$
1, 6, 11, 16......Find the 18th term of this A.P.
In a lottery there are 10 prizes and 25 blanks. What is the probability of getting a prize?
In a $\triangle\text{ABC},\text{AD}$ is a median and $\text{AL}\perp\text{BC}.$
Prove that:
→Prove that:
- $\text{AC}^2=\text{AD}^2+\text{BC}.\text{DL}+\Big(\frac{\text{BC}}{2}\Big)^2$
- $\text{AB}^2=\text{AD}^2-\text{BC}.\text{DL}+\Big(\frac{\text{BC}}{2}\Big)^2$
- $\text{AC}^2+\text{AB}^2=2\text{AD}^2+\frac{\text{1}}{2}\text{BC}^2$