Question

Sec(a-cos a ).(cot A+tan A )=secA.tan A

Answer 1

ANSWER

$(secA \: - \: cosA)(cotA \: + \: tanA) \\ \\ = > \: ( \frac{1}{cosA \: } \: - \: cosA \: )( \: \frac{cosA \: }{sinA \: } \: + \: \frac{sinA \: }{cosA \: } \: ) \\ \\ = > ( \frac{1 \: - \: {cos}^{2} }{cos} \: )( \frac{ {cos}^{2} A \: + \: {sin}^{2} A \: }{cosAsinA \: } \: ) \\ \\ = > \ \: ( \frac{ {sin}^{2} A \: }{cosA} \: ) ( \frac{1}{ cosAsinA \: }) \\ \\ = > \frac{ {sin}^{2}A }{ {cos}^{2}A \: sin A \: } \\ \\ = > \: \frac{sinA \: }{ {cos}^{2} A \: } \\ \\ = > \: ( \frac{1}{cosA \: } )( \frac{sinA \: }{cosA \: } ) \\ \\ = > secA \: . \: tanA \: = \: RHS \:$

{ HENCE PROVED }

INDENDITY USED :

➡ cos²A + sin²A = 1

➡ $\frac{1}{cosA} = secA$

➡ $\frac{sinA}{cosA} = tanA$

➡ $\frac{cosA}{sinA} = cotA$

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