Question

prove with process please​

Answer 1

1. Correct question:-

$(1-sinA)(1+cosecA)=cosA.cotA$

LHS:

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

$\bold{=(1-sinA)(\frac{sinA+1}{sinA})}\\\\=\bold{\frac{(1-sinA)(sinA+1)}{sinA} \\\\=\frac{(1)^2-sin^2A}{sinA} \\\\=\frac{cos^2A}{sinA} \\\\=\frac{cosA}{sinA} *cosA\\\\=cotA.cosA=RHS}$

2. LHS:

$=\bold{\frac{(cosA+sinA)(cos^2+sin^2a-sinA.cosA)}{cosA+sinA} +\frac{(cosA-sinA)(sin^2A+cos^2A+sinA.cosA)}{cosA-sinA} }$

$=\bold{cos^2A+sin^2A-sinA.cosA+sin^2A+cos^2A-sinA.cosA}\\=\bold{1+1}\\=\bold{2=RHS}$

Answer 2

Answer:

. Correct question:-

(1-sinA)(1+cosecA)=cosA.cotA(1−sinA)(1+cosecA)=cosA.cotA

LHS:

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

sinA

1

)

\begin{gathered}\bold{=(1-sinA)(\frac{sinA+1}{sinA})}\\\\=\bold{\frac{(1-sinA)(sinA+1)}{sinA} \\\\=\frac{(1)^2-sin^2A}{sinA} \\\\=\frac{cos^2A}{sinA} \\\\=\frac{cosA}{sinA} *cosA\\\\=cotA.cosA=RHS}\end{gathered}

2. LHS:

=\bold{\frac{(cosA+sinA)(cos^2+sin^2a-sinA.cosA)}{cosA+sinA} +\frac{(cosA-sinA)(sin^2A+cos^2A+sinA.cosA)}{cosA-sinA} }=

cosA+sinA

(cosA+sinA)(cos

2

+sin

2

a−sinA.cosA)

+

cosA−sinA

(cosA−sinA)(sin

2

A+cos

2

A+sinA.cosA)

\begin{gathered}=\bold{cos^2A+sin^2A-sinA.cosA+sin^2A+cos^2A-sinA.cosA}\\=\bold{1+1}\\=\bold{2=RHS}\end{gathered}

=cos

2

A+sin

2

A−sinA.cosA+sin

2

A+cos

2

A−sinA.cosA

=1+1

=2=RHS

Step-by-step explanation:

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