Question

(1+tanA+secA)(1+cotA-cosecA) Evaluate the following

Answer 1

GIVEN :

The expression is $(1+tanA+secA)(1+cotA-cosecA)$

TO EVALUATE :

The given expression $(1+tanA+secA)(1+cotA-cosecA)$

SOLUTION :

Given expression is $(1+tanA+secA)(1+cotA-cosecA)$

Now solving the given expression as below

$(1+tanA+secA)(1+cotA-cosecA)$

$=(1+\frac{sinA}{cosA}+\frac{1}{cosA})(1+\frac{cosA}{sinA}-\frac{1}{sinA})$

By using the trignometric formulae :

i) $tanx=\frac{sinx}{cosx}$

ii) $secx=\frac{1}{cosx}$

iii) $cotx=\frac{cosx}{sinx}$

iv) $cosecx=\frac{1}{secx}$

$=(\frac{(cosA+sinA)+1}{cosA})(\frac{(sinA+cosA)-1}{sinA})$

$=(\frac{(cosA+sinA)+1}{cosA})(\frac{(cosA+sinA)-1}{sinA})$

By using the Algebraic identity :

$(a+b)(a-b)=a^2-b^2$

$=\frac{(cosA+sinA)^2-1^2}{cosA sinA}$

By using the Algebraic identity :

$(a+b)^2=a^2+2ab+b^2$

$=\frac{cos^2A+sin^2A+2cosA sinA-1}{cosA sinA}$

By using the trignometric identity :

$cos^2x+sin^2x=1$

$=\frac{1+2cosA sinA-1}{cosA sinA}$

$=\frac{2cosA sinA}{cosA sinA}$

= 2

∴ $(1+tanA+secA)(1+cotA-cosecA)=2$

∴ the evaluated value for the given expression $(1+tanA+secA)(1+cotA-cosecA)$ is 2

Answer 2

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