Math, asked by chandrikasingh2036, 1 month ago

tan^2A-sin^2A=sin^2×tan^2A​

Answers

Answered by KuhuChoudhary
13

Answer

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

2

A−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

A

sin

2

A

−sin

2

A

=

cos

2

A

(sin

2

A−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

A

1−cos

2

A

=

cos

2

A

sin

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

2

A

Hence proved.

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