Question

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Answer 1

$\huge\color{green}\mathtt{Answer}$=

LHS = 1-cot(a)/sec(a)

RHS = sin2(a)/1-cos(a)

So,1+cotA/secA

=1+(cosA/sinA)/(1/cosA)

=1+cos²A/sinA

=(sinA+cos²A)/sinA

sin²A/(1-cosA)

=(1-cos²A)/(1-cosA)

=(1+cosA)(1-cosA)/(1-cosA)

=1+cosA

∴, LHS≠RHS

But, if LHS

=1+cotA/cosecA

=1+(cosA/sinA)/(1/sinA)

=1+cosA

Thus, LHS=RHS

$<marquee>Hope it helps you and don't forget to BRAINLIEST me...✌️$

Answer 2

Answer:

LHS = 1-cot(a)/sec(a)

RHS = sin2(a)/1-cos(a)

So,1+cotA/secA

=1+(cosA/sinA)/(1/cosA)

=1+cos²A/sinA

=(sinA+cos²A)/sinA

sin²A/(1-cosA)

=(1-cos²A)/(1-cosA)

=(1+cosA)(1-cosA)/(1-cosA)

=1+cosA

LHS≠RHS

But, if LHS=1+cotA/cosecA

=1+(cosA/sinA)/(1/sinA)

=1+cosA

HENCE, LHS=RHS