Question

prove that: sin2A +sin2A. tan2A =tan2A​

Answer 1

Answer:

Hence proved that \tan ^ { 2 } A - \sin ^ { 2 } A = \tan ^ { 2 } A \sin ^ { 2 } Atan 2

A−sin 2 A=tan 2 Asin 2 A

To prove:

\tan ^ { 2 } A - \sin ^ { 2 } A = \tan ^ { 2 } A \sin ^ { 2 } Atan 2

A−sin 2 A=tan 2 Asin 2 A

Solution:

Given,

\begin{gathered}\tan ^ { 2 } A - \sin ^ { 2 } A \\\end{gathered}

tan 2A−sin 2 A

We know that the value of \tan \theta = \frac { \sin \theta } { \cos \theta }tanθ=

cosθ

sinθ

\begin{gathered}\begin{array} { c } { \therefore \tan ^ { 2 } A - \sin ^ { 2 } A = \frac { \sin ^ { 2 } A } { \cos ^ { 2 } A } - \sin ^ { 2 } A } \\\\ { = \frac { \left( \sin ^ { 2 } A - \cos ^ { 2 } A \sin ^ { 2 } A \right) } { \cos ^ { 2 } A } } \end{array}\end{gathered}

∴tan 2 A−sin 2

A= cos 2 Asin 2 A −sin 2 A

= cos 2 A

(sin 2A−cos 2 Asin 2 A)

\begin{gathered}\begin{array} { l } { = \sin ^ { 2 } A \frac { 1 - \cos ^ { 2 } A } { \cos ^ { 2 } A } } \\\\ { = \frac { \sin ^ { 2 } A } { \cos ^ { 2 } A } \left( 1 - \cos ^ { 2 } A \right) } \\\\ { = \tan ^ { 2 } A \left( 1 - \cos ^ { 2 } A \right) } \end{array}\end{gathered}

=sin 2 A cos 2 A1−cos 2A

= cos 2Asin 2 A

(1−cos 2 A)

=tan 2 A(1−cos 2 A)

Since we know that the value of \left( 1 - \cos ^ { 2 } A \right) \text { is } \sin ^ { 2 } A(1−cos 2 A) is sin 2 A

Substituting this in the above terms, we get the following,

\tan ^ { 2 } A - \sin ^ { 2 } A = \tan ^ { 2 } A \sin ^ { 2 } Atan 2 A−sin 2 A=tan 2 Asin 2 A

Hence proved that the subtraction of \tan ^ { 2 } A \text { and } \sin ^ { 2 } Atan 2 A and sin 2 A will lead to the product of \tan ^ { 2 } A \text { and } \sin ^ { 2 } Atan 2 A and sin 2 A

Thus, \tan ^ { 2 } A - \sin ^ { 2 } A = \tan ^ { 2 } A \sin ^ { 2 } Atan 2 A−sin 2 A=tan 2 Asin 2A

Hence proved.

Answer 2

Step-by-step explanation:

LHS= sin²A+sin²A.tan²A

= sin²A+sin²A.sin²A/cos²A

= (sin²A.cos²A+sin⁴A)/cos²A

= sin²A(cos²A+sin²A)/cos²A since sin²A+cos²A=1

= sin²A/cos²A

= tan²A

= RHS