Question
If $A=30^{\circ}$ and $B=60^{\circ}$, verify that: $\frac{\sin (A-B)}{\sin A \cdot \sin B}=\cot B-\cot A$
✓
Answer
$A = 30^\circ$ and $B = 60^\circ$
$\text{L.H.S.}$
$=\frac{\sin (A-B)}{\sin A \cdot \sin B}$
$=\frac{\sin \left(30^{\circ}-60^{\circ}\right)}{\sin 30^{\circ} \times \sin 60^{\circ}}$
$=\frac{-\sin 30^{\circ}}{\sin 30^{\circ} \times \sin 60^{\circ}}$
$=\frac{-\frac{1}{2}}{\frac{1}{2} \times \frac{\sqrt{3}}{2}}$
$=-\frac{2}{\sqrt{3}}$
$\text{R.H.S.}$
$= \cot B - \cot A$
$= \cot 60^\circ -\cot 30^\circ$
$=\frac{1}{\sqrt{3}}-\sqrt{3}$
$=\frac{1-3}{\sqrt{3}}$
$=-\frac{2}{\sqrt{3}}$
$\Rightarrow \frac{\sin (A-B)}{\sin A \cdot \sin B}=\cot B-\cot A$
$\text{L.H.S.}$
$=\frac{\sin (A-B)}{\sin A \cdot \sin B}$
$=\frac{\sin \left(30^{\circ}-60^{\circ}\right)}{\sin 30^{\circ} \times \sin 60^{\circ}}$
$=\frac{-\sin 30^{\circ}}{\sin 30^{\circ} \times \sin 60^{\circ}}$
$=\frac{-\frac{1}{2}}{\frac{1}{2} \times \frac{\sqrt{3}}{2}}$
$=-\frac{2}{\sqrt{3}}$
$\text{R.H.S.}$
$= \cot B - \cot A$
$= \cot 60^\circ -\cot 30^\circ$
$=\frac{1}{\sqrt{3}}-\sqrt{3}$
$=\frac{1-3}{\sqrt{3}}$
$=-\frac{2}{\sqrt{3}}$
$\Rightarrow \frac{\sin (A-B)}{\sin A \cdot \sin B}=\cot B-\cot A$
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
A shopkeeper allows a discount of $12.5\%$ on the marked price and makes a profit of $20\%$. If the cost price is $Rs.4200,$ what should be the marked price?
→Find the loss percentage incurred by selling a watch for $Rs.1610$ that was purchased for $Rs.1750.$
→
The shape of the top of a table in a restaurant is that of a segment of a circle with centre O and $\angle$BOD = $90^{\circ}$. BO = OD = 60 cm.
Find : (i) the area of the top of the table;
(ii) the perimeter of the table. [Take $\pi$ = 3.14]
A 90% acid solution is mixed with a 97% acid solution to obtain 21 litres of a 95% solution. Find the quantity of each of the solutions to get the resultant mixture.
[Hint. Let x litres of 90% solution be mixed with y litres of 97% acid solution.
Then, x + y = 21 and 90% of x + 97% of y = 95% of 21.]
[Hint. Let x litres of 90% solution be mixed with y litres of 97% acid solution.
Then, x + y = 21 and 90% of x + 97% of y = 95% of 21.]
In the given figure: $A B \| F D, A C \| G E$ and $B D=C E$;
prove that: $(i)\mathrm{BG}=\mathrm{DF} \text { (ii) } \mathrm{CF}=\mathrm{EG}$
→prove that: $(i)\mathrm{BG}=\mathrm{DF} \text { (ii) } \mathrm{CF}=\mathrm{EG}$
In the adjoining figure, $\mathrm{OX}$ and $\mathrm{RX}$ are the bisectors of the angles $Q$ and $R$ respectively of the $\triangle P Q R$.If $X S \perp Q R$ and $X T \perp P Q$;
prove that: $(i) \triangle \mathrm{XTQ} \cong \triangle \mathrm{XSQ}.(ii) \mathrm{PX}$ bisects angle $\mathrm{P}$
→prove that: $(i) \triangle \mathrm{XTQ} \cong \triangle \mathrm{XSQ}.(ii) \mathrm{PX}$ bisects angle $\mathrm{P}$
In the given figure; $A B=B C$ and $A D=E C$. Prove that: $BD = BE$.
→Draw parallelogram $\text{ABCD}$ with the following data:$A B=6 \ cm , A D=5 \ cm$ and $\angle D A B=45^{\circ}$.Let $A C$ and $D B$ meet in $O$ and let $E$ be the mid$-$point of $B C$. Join $O E$.Prove that:$(i) OE \| AB;(ii) OE =\frac{1}{2} AB$.
→If $\tan \theta=\frac{5}{12}$ and $\theta$ is acute, find the values of sin $\theta$ and cos $\theta$.
→If $a-\frac{1}{a}=8$ and $a \neq 0$ find:$(i)\ a+\frac{1}{a};(ii)\ a^2-\frac{1}{a^2}$
→