Question

${\sec }^{6}A-{\tan }^{6}A=1+3{\tan }^{2}A+3{\tan }^{4}A$

Answer 1

Given$LHS=se{c}^{6}A-ta{n}^{6}A$.

W.K.T$a^3-b^3=(a-b)(a^2+ab+b^2)$

$=(se{c}^{2}A-ta{n}^{2}A)\left(\right(se{c}^{2}A{)}^2+se{c}^{2}Ata{n}^{2}A+ta{n}^{4}A)$

W.K.T$se{c}^2\theta =1+ta{n}^2\theta \left(or\right)se{c}^2\theta -ta{n}^2\theta =1.$

$=\left(1\right)\left(\right(se{c}^{2}A{)}^2+(1+ta{n}^{2}A)\left(ta{n}^{2}A\right)+\left(ta{n}^{4}A\right)$

$=(se{c}^{2}A{)}^2+(1+ta{n}^{2}A\left)\right(ta{n}^{2}A)+(ta{n}^{4}A)$

$=(1+ta{n}^{2}A{)}^2+(1+ta{n}^{2}A\left)\right(ta{n}^{2}A)+(ta{n}^{4}A)$

$=(1+ta{n}^{2}A)\left[\right(1+ta{n}^{2}A)+ta{n}^{2}A]+ta{n}^{4}A$

$=(1+ta{n}^{2}A)(1+2ta{n}^{2}A)+ta{n}^{4}A$

$=1+2ta{n}^{2}A+ta{n}^{2}A+2ta{n}^{4}A+ta{n}^{4}A$

$=1+3ta{n}^{2}A+3ta{n}^{4}A$

LHS=RHS.