Question

Please solve question no. 7. Please guys today is my test

Answer 1

here is your answer....

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Answer 2

Hey there !!


▶ Question :-


$If \sqrt{3} tan \theta = 3sin \theta , \: prove \: that( {sin}^{2} \theta - {cos}^{2} \theta) = \frac{1}{3} .$


▶ Solution :-

$\sqrt{3} tan \theta = 3sin \theta. \\ \\ = > \frac{tan \theta}{sin \theta} = \frac{3}{ \sqrt{3} } . \\ \\ = > \frac{ \frac{ \cancel {sin \theta}}{cos \theta} }{ \cancel{sin \theta}} = \frac{ \cancel{\sqrt{3}} \times \sqrt{3} }{ \cancel{\sqrt{3}} } . \\ \\ = > \frac{1}{cos \theta} = \sqrt{3} . \\ \\ = > cos \theta = \frac{1}{ \sqrt{3} } .$


▶ Now,


$\bf{Solving \: LHS } \\ \\ = {sin}^{2} \theta - {cos}^{2} \theta. \\ \\ = 1 - {cos}^{2} \theta - {cos}^{2} \theta. \\ ( \therefore {sin}^{2} \theta = 1 - {cos}^{2} \theta). \\ \\ = 1 - 2 {cos}^{2} \theta. \\ \\ = 1 - 2 {( \frac{1}{ \sqrt{3} } )}^{2} . \\ \\ = 1 - 2 \times \frac{1}{3} . \\ \\ = 1 - \frac{2}{3} . \\ \\ = \frac{3 - 2}{3} . \\ \\ \huge \boxed{ \boxed{ \bf = \frac{1}{3} .}} \checkmark \checkmark$


$\huge \bf \underline{ \mathbb{LHS = RHS .}}$



✔✔ Hence, it is proved ✅✅.

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THANKS


#BeBrainly.