tan^2A-sin^2A=sin^2×tan^2A
Answers
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.