Question

Tan2A - Sin2A = tan2A * sin2A. Prove

Answer 1

Answer:

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

To prove:

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

Solution:

Given,  

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

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

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

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

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

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

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

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

Thus, $\tan ^ { 2 } A - \sin ^ { 2 } A = \tan ^ { 2 } A \sin ^ { 2 } A$

Hence proved.

Answer 2

Answer:

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

Step-by-step explanation:

$LHS = tan^{2}A-sin^{2}A\\=\frac{sin^{2}A}{cos^{2}A}-sin^{2}A\\=sin^{2}A\big(\frac{1}{cos^{2}A}-1\big)\\=sin^{2}A\big(\frac{1-cos^{2}A}{cos^{2}A}\big)\\=\\=\frac{sin^{2}A}{cos^{2}A}\big(1-cos^{2}A)\big)\\=\frac{sin^{2}A}{cos^{2}A}\times sin^{2}A$

/* We know the Trigonometric identity:

1-cos²A=sin²A */

$=tan^{2}Asin^{2}A$

$=RHS$

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