Question

if tan theta = √3 find the value of sin 2theta-cos2theta


don't copy and paste i need solved answer​

Answer 1

Solution :

$\underline{\bf{Given\::}}}}$

tan Ф = √3

$\underline{\bf{Explanation\::}}}}$

Firstly, we attach a diagram of a right angle triangle, according to the question;

As we know that;

$\boxed{\bf{tan\:\theta= \frac{Perpendicular}{Base} }}}$

$\longrightarrow\sf{tan \:\theta = \dfrac{\sqrt{3} }{1} =\dfrac{BC}{AC} }$

$\underline{\boldsymbol{By\:using\:pythagoras\:theorem\::}}}$

$\mapsto\sf{(Hypotenuse)^{2} = (Base)^{2} + (Perpendicular)^{2} }\\\\\mapsto\sf{(AB)^{2} = (AC)^{2} + (BC)^{2}}\\\\\mapsto\sf{(AB)^{2} = (1)^{2} + (\sqrt{3} )^{2} } \\\\\mapsto\sf{(AB)^{2} = 1 + 3}\\\\\mapsto\sf{(AB)^{2} = 4}\\\\\mapsto\sf{AB= \sqrt{4} }\\\\\mapsto\bf{AB= 2\:unit}$

Now;

$\longrightarrow\sf{sin^{2} \theta - cos^{2} \theta}\\\\\longrightarrow\sf{\bigg(\dfrac{P}{H} \bigg)^{2} - \bigg(\dfrac{B}{H} \bigg)^{2}}\\\\\\\longrightarrow\sf{\bigg(\dfrac{\sqrt{3} }{2} \bigg)^{2} - \bigg(\dfrac{1}{2} \bigg)^{2} }\\\\\\\longrightarrow\sf{\dfrac{3}{4} - \dfrac{1}{4} }\\\\\\\longrightarrow\sf{\dfrac{3-1}{4} }\\\\\\\longrightarrow\sf{\cancel{\dfrac{2}{4} }}\\\\\longrightarrow\bf{1/2}$

Thus;

The value of sin²Ф - cos² Ф will be 1/2 .