Question

Prove that : sinA(1+tanA) +cosA(1+cotA) = secA + cosecA

Answer 1

answer starts from above ok

Answer 2

Answer:

$sinA(1+tanA)+cosA(1+cotA)=secA+cosecA$

Step-by-step explanation:

$LHS=sinA(1+tanA)+cosA(1+cotA)$

$=sinA[1+\frac{sinA}{cosA}]+cosA[1+\frac{cosA}{sinA}]$

$=sinA[\frac{cosA+sinA}{cosA}]+cosA[\frac{sinA+cosA}{sinA}$

$=\left(\frac{sinA}{cosA}\right)(sinA+cosA)+\left(\frac{cosA}{sinA}\right)(sinA+cosA)$

$=(sinA+cosA)[\frac{sinA}{cosA}+\frac{cosA}{sinA}]$

$=(sinA+cosA)[\frac{sin^{2}A+cos^{2}A}{sinAcosA}]$

$=(sinA+cosA)[\frac{1}{sinAcosA}]$

/*By Trigonometric identity:

sin²A+cos²A = 1 */

$=\frac{sinA}{sinAcosA}+\frac{cosA}{sinAcosA}$

$=\frac{1}{cosA}+\frac{1}{sinA}$

$=secA+cosecA$

$=RHS$

Therefore,

$sinA(1+tanA)+cosA(1+cotA)=secA+cosecA$

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