Question

(1 + tanA + secA)(1 + cotA - cosecA) = ?​

Answer 1

GIVEN :

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

TO EVALUATE :

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

SOLUTION :

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

Now solving the given expression as below

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

=(1+\frac{sinA}{cosA}+\frac{1}{cosA})(1+\frac{cosA}{sinA}-\frac{1}{sinA})=(1+cosAsinA+cosA1)(1+sinAcosA−sinA1)

By using the trignometric formulae :

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

ii) secx=\frac{1}{cosx}secx=cosx1

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

iv) cosecx=\frac{1}{secx}cosecx=secx1

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

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

By using the Algebraic identity :

(a+b)(a-b)=a^2-b^2(a+b)(a−b)=a2−b2

=\frac{(cosA+sinA)^2-1^2}{cosA sinA}=cosAsinA(cosA+sinA)2−12

By using the Algebraic identity :

(a+b)^2=a^2+2ab+b^2(a+b)2=a2+2ab+b2

=\frac{cos^2A+sin^2A+2cosA sinA-1}{cosA sinA}=cosAsinAcos2A+sin2A+2cosAsinA−1

By using the trignometric identity :

cos^2x+sin^2x=1cos2x+sin2x=1

=\frac{1+2cosA sinA-1}{cosA sinA}=cosAsinA1+2cosAsinA−1

=\frac{2cosA sinA}{cosA sinA}=cosAsinA2cosAsinA

= 2

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