Question

sec4 A-sec2 A=tan2 A+tan4A provide​

Answer 1

Step-by-step explanation:

LHS : sec⁴A - sec²A

=> sec²A(sec²A - 1)

=> (tan²A +1)(tan²A). [ tan²A + 1 = sec²A ]

=> tan⁴A + tan²A

RHS : tan⁴A + tan²A

=> LHS = RHS

HENCE PROVED

Answer 2

Appropriate Question:

• Prove that $\sf sec^4 \; A - sec^2 \; A = tan^2\; A + tan^4 \; A$

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$\underline{\bf{\dag} \:\mathfrak{Taking\; LHS: :}}$⠀⠀⠀

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$:\implies\sf \sf sec^4 \; A - sec^2 \\\\\\:\implies\sf sec^2 A \Big(sec^2 A - 1 \Big) \\\\\\:\implies\sf \Big(1 + tan^2 \; A\Big)\; \Big(tan^2\; A \Big) \\\\\\:\implies\sf tan^2 \; A + tan^4\:A \\\\\\:\implies\bf\pink{tan^2 \; A + tan^4\:A = RHS}$

⠀⠀⠀$\therefore$ LHS = RHS; Hence Proved!

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$\bigstar\:\sf Trigonometric\:Values :\\\begin{tabular}{|c|c|c|c|c|c|}\cline{1-6}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} &a \dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & \dfrac{\sqrt{3}}{2}& \dfrac{1}{\sqrt{2}}& \dfrac{1}{2}& 0\\\cline{1-6}Tan \theta& 0& \dfrac{1}{\sqrt{3}}& 1& \sqrt{3}& Not D \hat{e} fined \\\cline{1-6}\end{tabular}$