Question

prove that tan square a minus sin square equal to tan square a into sin square​

Answer 1

Step-by-step explanation:

${ \tan }^{2} a - { \sin }^{2} a = { \tan }^{2} a \times { \sin }^{2}a \\ \frac{ { \sin}^{2}a }{ { \cos }^{2} a} - { \sin }^{2} a = { \tan }^{2} a \times { \sin }^{2}a \: \\ \\ \frac{ { \sin }^{2}a(1 - { \cos }^{2}a) }{ { \cos}^{2}a } = { \tan }^{2} a \times { \sin }^{2}a$

${ \ \tan }^{2}a \times { \sin}^{2}a = { \ \tan }^{2}a \times { \sin}^{2}a \:$

Identity used:

${ \sin}^{2} a + { \cos }^{2} a = 1$

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Answer 2

tep-by-step explanation:

Consider the provided information.

\sin^2A\cos^2B-\cos^2A\sin^2B=\sin^2A-\sin^2B

Consider the LHS.

\sin^2A\cos^2B-\cos^2A\sin^2B

\sin^2A(1-\sin^2B)-(1-\sin^2A)\sin^2B               (∴\cos^2x=1-\sin^2x)

\sin^2A-\sin^2A\sin^2B-\sin^2B+\sin^2A\sin^2B

\sin^2A-\sin^2B

Hence, proved.

Step-by-step explanation: