Question

Prove that: tan^4 A + sec^4 A = 1 + 2tan^2 A / cos^2 A

Answer 1

$\large\underline{\sf{Solution-}}$

Consider,

$\sf \: {tan}^{4}A + {sec}^{4}A$

can be rewritten as

$\sf \: = \: {sec}^{4}A + {tan}^{4}A$

$\sf \: = \: {( {sec}^{2}A) }^{2}+ {( {tan}^{2}A) }^{2}$

We know,

$\boxed{\sf \: {x}^{2} + {y}^{2} = {(x - y)}^{2} + 2xy \: }$

So, using this algebraic identity, we get

$\sf \: = \: {( {sec}^{2}A - {tan}^{2}A ) }^{2} \: + \: 2\: {sec}^{2}A \: {tan}^{2}A$

We know,

$\boxed{\begin{aligned}& \qquad \:\sf \: {sec}^{2}A - {tan}^{2}A = 1 \qquad \: \\ \\& \qquad \:\sf \: secA= \dfrac{1}{cosA} \end{aligned}} \qquad$

So, using these Identities, we get

$\sf \: = \: {(1) }^{2}+ 2\times \dfrac{1}{ {cos}^{2} A} \times {tan}^{2}A$

$\sf \: = \: 1+ \dfrac{2 {tan}^{2} A}{ {cos}^{2} A}$

Hence,

$\implies\sf \: \boxed{\bf \: {tan}^{4}A + {sec}^{4}A = 1+ \dfrac{2 {tan}^{2} A}{ {cos}^{2} A} \: }$

$\rule{190pt}{2pt}$

Additional Information

$\begin{gathered}\: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sinx = \dfrac{1}{cosecx} }\\ \\ \bigstar \: \bf{cosx = \dfrac{1}{secx} }\\ \\ \bigstar \: \bf{tanx = \dfrac{sinx}{cosx} = \dfrac{1}{cotx} }\\ \\ \bigstar \: \bf{cot x= \dfrac{cosx}{sinx} = \dfrac{1}{tanx} }\\ \\ \bigstar \: \bf{cosec x = \dfrac{1}{sinx} }\\ \\ \bigstar \: \bf{secx = \dfrac{1}{cosx} }\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 } \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\\\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Step-by-step explanation:

Hope this helps you!!!!!