Question

Explain trigonometry identities

Answer 1

A trigonometric identity is an equation involving trigonometric ratios of an angle, where the equation holds true for a defined range of values of the angle.
Let us take any acute angle taking point P on OB we draw perpendicular PQ on OAB. Thus we have a right angled triangle OPQ. If we suppose \angle $POQ=\theta$ then


$sin\theta =\frac{PQ}{OP},cos\theta =\frac{OQ}{OP}tan\theta =\frac{PQ}{OP}sec\theta =\frac{OP}{PQ}=\frac{1}{cos\theta }cosec\theta =\frac{OP}{PQ}=\frac{1}{sin\theta }cot\theta =\frac{OP}{PQ}=\frac{1}{tan\theta }tan\theta =\frac{PQ}{OQ}=\frac{PQ}{OP}÷\frac{OP}{OP}$
$=\frac{sin\theta }{cos\theta }cot\theta =\frac{cos\theta }{sin\theta }In\Delta POQ,\angle Q={90}^{\circ }$
$\therefore$ By pythagoras theorem
we have $O{Q}^2+P{Q}^2=O{P}^2$ .....(1)
Dividing both sides (i) by OP^2 we get
$\frac{O{Q}^2}{O{P}^2}+\frac{P{Q}^2}{O{P}^2}=\frac{O{P}^2}{O{P}^2}{\left(\frac{OQ}{OP}\right)}^2+{\left(\frac{PQ}{OP}\right)}^2=1$
or, $(cos\theta {)}^2+(sin{\theta }^2)=1$
or $co{s}^2\theta +si{n}^2\theta =1$ -----(2)
Similarly, dividing both sides of (1) by $O{Q}^2$ We get
$\frac{O{Q}^2}{O{Q}^2}+\frac{P{Q}^2}{O{Q}^2}=\frac{O{P}^2}{O{Q}^2}$
or, $1+{\left(\frac{PQ}{OQ}\right)}^2+{\left(\frac{OP}{OQ}\right)}^2$
Or, $1+(tan\theta {)}^2=(sec\theta {)}^2$
or, $1+ta{n}^3\theta =se{c}^2\theta$ -----(3)
A gain dividing both sides of (i) by $P{Q}^2$, we get
${\left(\frac{OQ}{PQ}\right)}^2+1={\left(\frac{OP}{PQ}\right)}^2$
Or, $(cot\theta {)}^2+1=(cosec\theta {)}^2$
or, $1+co{t}^2\theta =cose{c}^2\theta$ ----(4)
Here, (2), (3) and (4) are all identities, these can be used to express on e trigonometric ratio to other trogonometric ratio. For example
$co{s}^2\theta +si{n}^2\theta =1si{n}^2\theta =1-co{s}^2\theta sin\theta =\pm \sqrt{1-co{s}^2\theta }\therefore sin\theta =\sqrt{1-co{s}^2\theta }$
similarly.
$cos\theta =\sqrt{1-si{n}^2\theta },sec\theta =\sqrt{1+ta{n}^2\theta }tan\theta =\sqrt{se{c}^2\theta -1},cosec\theta =\sqrt{1+co{t}^2\theta }cot\theta =\sqrt{cose{c}^2\theta -1},for{o}^{\circ }<\theta <{90}^{\circ }$