Question

1/1+sin²A + 1/cos²A + 1/1+sec²A + 1/1+cosec²A = 2​

Answer 1

Answer:

ago

(tanA +sinA)/(tanA–sinA)=(secA+1)/(secA–1)

Then we taking LHS to prove it

By rationalize of denominator (tanA–sinA)we get

»(tanA+sinA)(tanA+sinA)/(tanA-sinA)(tanA+sinA)

»(tan²A+2sinA.tanA+sin²A)/(tan²A–sin²A)

Because,(a+b)²=a²+2a.b+b² &(a²–b²)=(a–b)(a+b)

»{(sec²A–1)+2sin²A.secA +sin²A}/sin²A(sec²A–1)

Because tan²A=(sec²A–1) & tanA =sinA.secA

»{(sec²A-1)+sin²A(2secA+1)}/sin²A(sec²A–1)

»1/sin²A + (2secA+1)/(sec²A–1)

»1/(1–cos²A) + (2secA+1)/(sec²A–1)

Because sin²A=(1–cos²A) and cosA =1/secA

»1/(1 –1/sec²A) + (2secA+1)/(sec²A–1)

»sec²A/(sec²A–1) + (2secA+1)/(sec²A–1)

»(sec²A + 2secA +1)/(sec²A–1)

»(secA+1)²/(secA–1)(secA+1)

(secA+1)/(secA–1),proved

Here, LHS=RHS