Question

sec
(1-sin
)(sec 0 + tan 6)=1.​

Answer 1

To Prove

$\mathsf{secA(1 - sinA)(secA + tanA) = 1}$

Solution

$\mathsf{\implies\:secA(1 - sinA)(secA + tanA) = 1}$

$\mathsf{\implies\:\frac{1}{cosA}(1 - sinA)(\frac{1}{cosA} + \frac{sinA}{cosA})}$

$\mathsf{\implies\:(\frac{1}{cosA} - \frac{sinA}{cosA})(\frac{1}{cosA} + \frac{sinA}{cosA})}$

$\fbox{\mathsf{\red{Using\:identity:\: (a-b) (a+b) = a^2 - b^2}}}$

$\mathsf{\implies\:(\frac{1}{cosA})^2 - (\frac{sinA}{cosA})^2 }$

$\mathsf{\implies\:\frac{1}{cos^2A} - \frac{sin^2A}{cos^2A} }$

Taking LCM

$\mathsf{\implies\:\frac{1 - sin^2A}{cos^2A} }$

$\mathsf{\implies\:\frac{\cancel{cos^2A}}{\cancel{cos^2A}} }$

$\mathsf{\implies\: 1}$

Hence Proved

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Trigonometric Identity used

• SecA = 1/ CosA
• tanA = SinA/ CosA
• Sin²A + Cos²A = 1

$\mathsf{\rightarrow}$1 - Sin²A = Cos²A

Answer 2

The question posted seems meaningless post it correctly