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Answer 1 · 2017-04-23T20:06:37

See below.

Starting from the LHS.

LHS = $\frac{3 - 4 {sin}^{2} A}{{cos}^{2} A}$ $(3 - 4sin^2A)/(cos^2A)$

$= \frac{3 - (3 {sin}^{2} A + {sin}^{2} A)}{{cos}^{2} A}$ $= [3 - (3sin^2A + sin^2A)]/(cos^2A)$

$= \frac{3 - 3 {sin}^{2} A - {sin}^{2} A}{{cos}^{2} A}$ $= (3-3sin^2A - sin^2A)/(cos^2A)$

$= \frac{3 (1 - {sin}^{2} A) - {sin}^{2} A}{{cos}^{2} A}$ $= (3(1-sin^2A) - sin^2A)/(cos^2A)$

$= \frac{3 {cos}^{2} A - {sin}^{2} A}{{cos}^{2} A}$ $= (3cos^2A - sin^2A)/(cos^2A)$

$= \frac{3 {cos}^{2} A}{{cos}^{2} A} - \frac{{sin}^{2} A}{{cos}^{2} A}$ $= (3cos^2A)/(cos^2A) - (sin^2A)/(cos^2A)$

$= 3 - {tan}^{2} A$ $= 3-tan^2A$ = RHS

Answer 2 · 2017-04-24T09:36:51

LHS = $\frac{3 - 4 {sin}^{2} A}{{cos}^{2} A}$ $(3 - 4sin^2A)/(cos^2A)$

= $\frac{3}{{cos}^{2} A} - \frac{4 {sin}^{2} A}{{cos}^{2} A}$ $3/cos^2A - (4sin^2A)/cos^2A$

= $3 {sec}^{2} A - 4 {tan}^{2} A$ $3sec^2A - 4tan^2A$

= $3 (1 + {tan}^{2} A) - 4 {tan}^{2} A$ $3(1+tan^2A )- 4tan^2A$

= $3 + 3 {tan}^{2} A - 4 {tan}^{2} A$ $3+3tan^2A - 4tan^2A$

= $3 - {tan}^{2} A = R H S$ $3 - tan^2A=RHS$