Question

please ans trignometry class 12th​

Answer 1

proof:-Divide numerator and denominator by cos∅

on Dividing we will get

LHS={(sec∅-tan∅)}^1/2/{(sec∅+tan∅)}1/2

multiplying by sec∅-tam∅ in both we get

LHS={sec∅-tan∅)^2}1/2/(1)^1/2

LHS=sec∅-tan∅

the value of sec∅-tan∅, should always be positive because we had taken it out from square root by squaring it

and square of any no. is always positive...

the value of sec∅ is greater than tan∅ in ∅ between π/2 to -π/2 ,,

so,sec∅-tan∅ will be positive

but the value of tan∅ become greater in ii) and third quadrant i.e from ∅ between π/2 to 3π/2

sec∅-tan∅, become negative in this case,whic is not possible

so to make it positive we multiply it by negative sign

then the value will be -sec∅+tan∅

{hope it helps}

Answer 2

Given:

√(1-sinA)/(1+SinA)

Solution:

Take L.H.S

→√(1-sinA)/(1+SinA)

Rationalise the given equation,

→√[(1-SinA)/(1+sinA)]×[(1-sinA)/(1+sinA)

→√(1-sin²A)/(1-sin²A)

→√(1-sin²A)/cos²A

→(1-sinA)/cosA

→1/cosA - sinA/cosA

→secA-tanA

mul by (-1)

→-secA+tan