Question

sec4a(1_sin4a) _2tan2A=1​

Answer 1

QUESTION :-

$\leadsto \tt \: sec {}^{4}A(1 - sin {}^{4}A) - 2tan {}^{2}A = 1$

SOLUTION :-

$\leadsto \tt { \underline{L. H. S: \: }}\: \: sec {}^{4}A(1 - sin {}^{4}A) - 2tan {}^{2}A \\ \\ \leadsto \tt \: sec {}^{4}A - sec {}^{4}A.sin {}^{4}A - 2tanA \\ \\ \leadsto \tt \: \frac{1}{cos{}^{4}A } - \frac{1}{cos {}^{4}A} .sin {}^{4}A \: - 2 \frac{sin {}^{2}A }{cos {}^{2} A} \\ \\ \leadsto \tt \: \frac{1}{cos{}^{4}A } - \frac{sin {4}^{2}A }{cos {}^{4}A} \: - 2 \frac{sin {}^{2}A }{cos {}^{2} A} \\ \\ \leadsto \tt \: \frac{(1 - sin {}^{4}A) }{cos {}^{2}A } - \frac{2sin {}^{2}A }{cos {}^{2}A } \\ \\ \leadsto \tt \: \frac{[1 - (sin {}^{2} A) {}^{2}] }{cos {}^{2}A } - \frac{2sin {}^{2}A }{cos {}^{2}A } \: \\ \\ \leadsto \tt \: \frac{(1 + sin {}^{2} A) \: (1 - sin {}^{2}A) }{cos {}^{4}A } - \frac{2sin {}^{2}A }{cos {}^{2}A } \: \\ \\ \leadsto \tt \: \frac{(1 + sin { }^{2} A) \times cos {}^{2} A}{cos {}^{4}A } - \frac{2sin {}^{4}A }{cos {}^{4}A } \\ \\ \leadsto \tt \: \frac{1 + sin {}^{2} A}{cos {}^{2}A } - \frac{2sin {}^{2} }{cos {}^{2}A } \\ \\ \leadsto \tt \: \frac{1 + sin {}^{2}A - 2sin {}^{2}A }{cos {}^{2}A } \\ \\ \leadsto \tt \: \frac{1 - sin {}^{2}A }{cos {}^{2}A } \\ \\ \leadsto \tt \: \frac{ \cancel{cos {}^{2}A }}{ \cancel{cos {}^{2}A } } \\ \\ \leadsto \tt 1 \\ \\ \leadsto \tt \: { \underline{R. H. S: }} \: \: \\ \\ \leadsto \tt \: { \boxed{ \boxed{ \tt{ \red{L. H. S = R. H. S \: }}}}} \: \\ \\ \leadsto \tt \:.°. \: { \green{ \underline{sec {}^{4}A(1 - sin {}^{4}A) - 2tan {}^{2}A = 1}}}$

Answer 2

Given LHS :

• $\sf{Sec^4\:A(1-sin^4\:A)-2\:tan^2\:A}$

To prove :

• LHS = 1.

Proof :

Taking LHS :

= $\sf{Sec^4\:A(1-sin^4\:A)-2\:tan^2\:A}$

Open the brackets,

= $\sf{Sec^4\:A\:-\:Sec^4\:A\:\times\:Sin^4\:A\:-\:2\:Tan^2\:A}$

= $\sf{\dfrac{1}{Cos^4\:A}\:-\:\dfrac{1}{Cos^4\:A}\:\times\:sin^4\:A\:-\:2\:\times\:\dfrac{Sin^2A}{Cos^2\:A}}$

$\sf{\red{\Big[\because\:Sec\:A\:=\:\dfrac{1}{Cos\:A}\:\:tan\:A\:=\:\dfrac{sin\:A}{Cos\:A}\Big]}}$

= $\sf{\dfrac{1}{cos^4\:A}\:-\:\dfrac{Sin^4\:A}{Cos^4\:A}\:-\:\dfrac{2\:sin^2\:A}{Cos^2\:A}}$

= $\sf{\dfrac{(1-sin^4\:A)}{Cos^4\:A}\:-\:\dfrac{2sin^2\:A}{Cos^2A}}$

= $\sf{\dfrac{\big[1+(sin^2A)^2\big]}{Cos^4A}\:-\:\dfrac{2sin^2A}{Cos^2\:A}}$

= $\sf{\dfrac{(1+sin^2\:A) (1-sin^2A)}{Cos^4A}\:-\:\dfrac{2sin^2A}{Cos^2A}}$

$\sf{\purple{\Big[a^2-b^2=(a+b)(a-b)\big]}}$

= $\sf{\dfrac{(1+sin^2A)\:\times\:\cancel{cos^2A}}{\cancel{Cos^4\:A}}\:-\:\dfrac{2sin^2A}{Cos^2A}}$

= $\sf{\dfrac{(1+sin^2A)}{Cos^2A}-\:\dfrac{2sin^2A}{Cos^2A}}$

= $\sf{\dfrac{1+sin^2\:-\:2\:sin^2A}{Cos^2A}}$

= $\sf{\dfrac{1-sin^2A}{Cos^2\:A}}$

= $\sf{\cancel{\dfrac{Cos^2\:A}{Cos^2\:A}}}$

$\sf{\green{\Big[\because\:1-sin^2A\:=\:Cos^2\:A\:\:;\because\:Cos^2A\:=\:1-sin^2A\Big]}}$

= $\sf{1}$

LHS = RHS.