1.

Prove that tan2A+cot2A+2=sec2A*cosec2A​

Answer»

Step-by-step explanation:

\\ \displaystyle \bf \large\blue{Question \:: - } \\

PROVE that :-

\displaystyle\sf {tan}^{2} A + {cot}^{2} A+ 2 = {sec}^{2} A + {cosec}^{2}

\\ \displaystyle \bf \large \red{Solution \: : - } \\

  • tan²A = sin²A/cos²A
  • cot²A = cos²A/sin²A

LHS :-

\displaystyle \sf {tan}^{2} A + {cot}^{2} A+ 2 \\ \\ \displaystyle\sf \: \dfrac{ {sin}^{2} A}{ {cos}^{2} A} + \dfrac{ {cos}^{2} A}{ {sin}^{2} A} + 2 \\ \\ \displaystyle \sf \: take \: lcm \: we \: get \: : - \\ \\ \displaystyle \sf \: \dfrac{ {sin}^{4} A + {cos}^{4} A + 2 {cos}^{2}A. {sin}^{2} A}{ {cos}^{2}A. {sin}^{2} A} \\

use IDENTITY a⁴ + b⁴ = ( a² + b² ) – 2a².b²

\\ \displaystyle \sf \: \dfrac{( { {sin}^{2}A + {cos}^{2}A) }^{2} - 2 {sin}^{2}A. {cos}^{2}A + 2 {cos}^{2}A. {sin}^{2}A}{ {cos}^{2} A. {sin}^{2}A } \\ \\ \displaystyle\sf \dfrac{( { {sin}^{2}A+ {cos}^{2}A) }^{2} \cancel{ - 2 {sin}^{2}A. {cos}^{2}A} \cancel{ + 2 {cos}^{2}A. {sin}^{2}A}}{ {cos}^{2} A. {sin}^{2}A} \\ \\ \displaystyle \sf \: \dfrac{ {( {sin}^{2}a + {cos}^{2}a) }^{2} }{ {cos}^{2}A. {sin}^{2} A} \\ \\ \displaystyle \sf \: \frac{ {(1)}^{2} }{ {cos}^{2} A + {sin}^{2} } \\ \\ \displaystyle \sf \frac{1}{ {cos}^{2} A} . \dfrac{1}{ {sin}^{2}A} \\ \\ \displaystyle \sf \: {sec}^{2}A. {cosec}^{2} A

HENCE,

\sf LHS \: = RHS


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