In $\triangle PQR ,$ right $-$ angled at $Q , PR + QR =25\, cm$ and $PQ =5\, cm .$ Determine the values of $\sin P, \cos P$ and $\tan P$.
Given that, $PR + QR =25$
$PQ =5$
Let $PR$ be $x$.
Therefore, $QR =25-x$
Applying Pythagoras theorem in $\triangle PQR$, we obtain
$PR ^{2}= PQ ^{2}+ QR ^{2}$
$x^{2}=(5)^{2}+(25-x)^{2}$
$x^{2}=25+625+x^{2}-50 x$
$50 x=650$
$x=13$
Therefore, $PR =13 \,cm$
$Q R=(25-13) \,cm =12\, cm$
$\sin P =\frac{\text { Side opposite to } \angle P }{\text { Hypotenuse }}=\frac{ QR }{ PR }=\frac{12}{13}$
$\cos P =\frac{\text { Side adjacent to } \angle P }{\text { Hypotenuse }}=\frac{ PQ }{ PR }=\frac{5}{13}$
$\tan P =\frac{\text { Side opposite to } \angle P }{\text { Side adjacent to } \angle P }=\frac{ QR }{ PQ }=\frac{12}{5}$
Prove that
$\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta},$ using the identity
$\sec ^{2} \theta=1+\tan ^{2} \theta$
If $A , B$ and $C$ are interior angles of a triangle $ABC ,$ then show that
$\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}$
Express $\cot 85^{\circ}+\cos 75^{\circ}$ in terms of trigonometric ratios of angles between $0^{\circ}$ and $45^{\circ}$
If $\tan ( A + B )=\sqrt{3}$ and $\tan ( A - B )=\frac{1}{\sqrt{3}} ; 0^{\circ}< A + B \leq 90^{\circ} ; A > B ,$ find $A$ and $B$
Consider $\triangle ACB$, right-angled at $C$, in which $AB =29$ units, $BC =21$ units and $\angle ABC =\theta$ (see $Fig.$). Determine the values of
$(i)$ $\cos ^{2} \theta+\sin ^{2} \theta$
$(ii)$ $\cos ^{2} \theta-\sin ^{2} \theta$