Question

please try to answer this question correctly​

Answer 1

Answer:

$8.Given \: \sqrt{3} tan\theta = 1$

$\implies tan\theta = tan \: 30 \degree$

$\implies \theta = 30\degree$

$\red {Value \: of \: sin2\theta - cos2\theta }$

$\begin{gathered} = sin 2\times 30 - cos 2\times 30 \\= sin 60 - cos 60\\= \frac{\sqrt{3}}{2} - \frac{1}{2}\\= \frac{ \sqrt{3}-1}{2}\end{gathered}$

Therefore.,

Value of sin2θ−cos2θ

=√3-1/2

9. We have 4 tan Ф = 3

$\longrightarrow\sf{tan\theta=\dfrac{3}{4}=\bigg[\dfrac{Perpendicular}{Base} \bigg]}$

By Using pythagoras theorem

$\begin{gathered}\longrightarrow\sf{(hypotenuse)^{2} =(base)^{2} +(perpendicular)^{2} }\\\\\longrightarrow\sf{(AC)^{2} =(BC)^{2} +(AB)^{2} }\\\\\longrightarrow\sf{(AC)^{2} =(4)^{2} +(3)^{2} }\\\\\longrightarrow\sf{(AC)^{2} =16+9}\\\\\longrightarrow\sf{(AC)^{2} =25}\\\\\longrightarrow\sf{AC=\sqrt{25} }\\\\\longrightarrow\sf{AC=5\:units}\end{gathered}$

Now;

$\begin{gathered}\longrightarrow\sf{sin\theta\times cos\theta=\dfrac{12}{25} }\\\\\\\longrightarrow\sf{\dfrac{Perpendicular}{Hypotenuse} \times \dfrac{Base}{Hypotenuse} =\dfrac{12}{25} }\\\\\\\longrightarrow\sf{\dfrac{AB}{AC} \times \dfrac{BC}{AC} =\frac{12}{25}}\\\\\\\longrightarrow\sf{\dfrac{3}{5} \times \dfrac{4}{5} =\dfrac{12}{25}} \\\\\\\longrightarrow\bf{\dfrac{12}{25} =\dfrac{12}{25} \:\:\:[L.H.S=R.H.S]}\end{gathered}$

[L.H.S=R.H.S]

Hence, Proved.

10-in image