Question

10. Prove that:
tan2A - sin2 A = tan²A * sin2A​

Answer 1

AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• Tan²A-sin²A=Tan²A×Sin²A

${\sf{\green{\underline{\large{To\:find}}}}}$

• LHS=RHS

${\sf{\pink{\underline{\Large{Explanation}}}}}$

• tan2A - sin2 A

Breaking tan²A -sin²A in (a+b)(a-b)

=>tan²A -sin²A

$\tt\implies\:Tan^2A-Sin^2A=(TanA-CosA)(Tan+CosA)\\{\implies}(\frac{SinA}{CosA}-CosA)(\frac{SinA}{CosA}+CosA)\\{\implies}\frac{Sin-sin\times\:cos}{cos}\frac{Sin+sin\times\:cos}{cos}\\{\implies}\frac{Sin(1-cos)}{cos}\frac{Sin(1+cos}{cos}$

=>sin²A/cos²A (1-cos²)

=>Tan²A×Sin²A

Answer 2

$\large \underline\textsf{Given:}$

$</p><p> \tan^2A - \sin^2A$

$\large\underline \textsf {To \: prove:}$

$</p><p> \tan^2A - \sin^2A =\tan^2A.\sin^2A$

$\large\underline\textsf{Solution:}$

We have,

$\tan {}^{2} A - \sin {}^{2} A$

$\text{Also, we \: know \: that} \: \tan\theta = \frac{\sin\theta}{\cos\theta} \\ \\ </p><p>\implies {\tan}^{2}A - {\sin}^{2}A = \frac{{ \sin }^{2}A}{ {\cos }^{2}A } - { \sin}^{2}A \: \\ \\ </p><p> = \frac{{ \sin}^{2}A-{ \sin}^{2}A.{ \cos}^{2}A}{{ \cos}^{2}A} \\ \\ </p><p>={\sin}^{2}A\left(\frac{1-{\cos}^{2}A}{{\cos}^{2}A}\right) \\ \\ </p><p>=\frac{{\sin}^{2}A}{{\cos}^{2}A}\left(1-{\cos}^{2}A\right) \\ \\ </p><p>\text{Since \: we \: know \: that} \\ \\ </p><p>\left[{\sin}^{2}\theta +{\cos}^{2}\theta=1 \right]\\\left [\implies{\sin}^{2}\theta=1-{\cos}^{2}\theta\right] \\ </p><p>\text{and}\\</p><p>\left[ \tan A = \frac{ \sin A}{ \cos A } \right]$

Substituting the value

We get,

$</p><p> \boxed{ \tan^2A - \sin^2A =\tan^2A.\sin^2A}$

$\textsf{ Hence, proved }$