Question

SinA/ secA +tanA-1 + cosA/ cosecA + cotA =1

Answer 1

$\huge\boxed{\fcolorbox{red}{yellow}{Shailesh}}$

SinA/secA+tanA-1+cosA/cosecA+cotA-1

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

=sinA/{(1+sinA-cosA)/cosA}+cosA/{(1+cosA-sinA)/sinA}

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

=sinAcosA[(1+cosA-sinA+1+sinA-cosA)/(1+sinA-cosA)(1+cosA-sinA)]

=2sinAcosA/(1+sinA-cosA+cosA+sinAcosA-cos²A-sinA-sin²A+sinAcosA)

=2sinAcosA/{1+2sinAcosA-(sin²A+cos²A)}

=2sinAcosA/(1+2sinAcosA-1)

=2sinAcosA/2sinAcosA

=1 (Proved)

$\bf\huge\underline\red{Follow me}$

Answer 2

Answer:

$\huge\star\mathfrak\blue{{Answer:-}}$

Step-by-step explanation:

SinA/secA+tanA-1+cosA/cosecA+cotA-1

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

=sinA/{(1+sinA-cosA)/cosA}+cosA/{(1+cosA-sinA)/sinA}

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

=sinAcosA[(1+cosA-sinA+1+sinA-cosA)/(1+sinA-cosA)(1+cosA-sinA)]

=2sinAcosA/(1+sinA-cosA+cosA+sinAcosA-cos²A-sinA-sin²A+sinAcosA)

=2sinAcosA/{1+2sinAcosA-(sin²A+cos²A)}

=2sinAcosA/(1+2sinAcosA-1)

=2sinAcosA/2sinAcosA

=1 (Proved)