In triangle $ABC ,$ right -angled at $B ,$ if $\tan A =\frac{1}{\sqrt{3}},$ find the value of:
$(i)$ $\sin A \cos C+\cos A \sin C$
$(ii)$ $\cos A \cos C-\sin A \sin C$
In triangle $ABC ,$ right -angled at $B ,$ if $\tan A =\frac{1}{\sqrt{3}},$ find the value of:
$(i)$ $\sin A \cos C+\cos A \sin C$
$(ii)$ $\cos A \cos C-\sin A \sin C$
$\tan A =\frac{1}{\sqrt{3}}$
$\frac{ BC }{ AB }=\frac{1}{\sqrt{3}}$
If $B C$ is $k$, then $A B$ will be $\sqrt{3} k,$ where $k$ is a positive integer.
$\ln \triangle ABC ,$
$-A C^{2}=A B^{2}+B C^{2}$
$(\sqrt{3} k)^{2}+(k)^{2}$
$=3 k^{2}+k^{2}=4 k^{2}$
$AC =2 k$
$\sin A=\frac{\text { Side opposite to } \angle A }{\text { Hypotenuse }}=\frac{ BC }{ AC }=\frac{k}{2 k}=\frac{1}{2}$
$\cos A=\frac{\text { Side adjacent to } \angle A }{\text { Hypotenuse }}=\frac{ AB }{ AC }=\frac{\sqrt{3} k}{2 k}=\frac{\sqrt{3}}{2}$
$\sin C=\frac{\text { Side opposite to } \angle C }{\text { Hypotenuse }}=\frac{ AB }{ AC }=\frac{\sqrt{3} k}{2 k}=\frac{\sqrt{3}}{2}$
$\cos C=\frac{\text { Side adjacent to } \angle C }{\text { Hypotenuse }}=\frac{ BC }{ AC }=\frac{k}{2 k}=\frac{1}{2}$
$(i)$ $\sin A \cos C+\cos A \sin C$
$=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)+\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)=\frac{1}{4}+\frac{3}{4}$
$=\frac{4}{4}=1$
$(ii)$ $\cos A \cos C-\sin A \sin C$
$=\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right)-\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}=0$
Similar Questions
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$(\operatorname{cosec} \theta-\cot \theta)^{2}=\frac{1-\cos \theta}{1+\cos \theta}$
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$(\operatorname{cosec} \theta-\cot \theta)^{2}=\frac{1-\cos \theta}{1+\cos \theta}$
If $\angle A$ and $\angle B$ are acute angles such that $\cos A =\cos B ,$ then show that $\angle A =\angle B$.
If $\angle A$ and $\angle B$ are acute angles such that $\cos A =\cos B ,$ then show that $\angle A =\angle B$.
If $A , B$ and $C$ are interior angles of a triangle $ABC ,$ then show that
$\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}$
If $A , B$ and $C$ are interior angles of a triangle $ABC ,$ then show that
$\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}$
Evaluate:
$\sin 25^{\circ} \cos 65^{\circ}+\cos 25^{\circ} \sin 65^{\circ}$
Evaluate:
$\sin 25^{\circ} \cos 65^{\circ}+\cos 25^{\circ} \sin 65^{\circ}$
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\operatorname{cosec} A+\cot A,$ using the identity $\operatorname{cosec}^{2} A=1+\cot ^{2} A$
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\operatorname{cosec} A+\cot A,$ using the identity $\operatorname{cosec}^{2} A=1+\cot ^{2} A$