Question

sin A/ sec A+ tan A -1 + cos A / cosec A+cot A+1

Answer 1

Step-by-step explanation:

sinA/secA+tanA-1 + cosA/

cosecA+cotA+1

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

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

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

=sinAcosA[2(1+sinA)/1+sin²A+2SinA-cos²A]

=sinAcosA[2(1+SinA)/1+sin²A+2SinA-1+sin²A]

=SinA cosA[2(1+sinA)/2sinA(1+sinA)]

=cosA

Answer 2

Step-by-step explanation:

$\frac{sinA}{secA+tanA-1 }+\frac{cosA} {cosecA+cotA+1}$

$=\frac{sinAcosA}{(1+sinA)-cosA}+\frac{cosAsinA}{1+cosA+sinA}\\=sinAcosA[\frac{1}{(1+sinA)-cosA}+\frac{1}{(1+sinA)+cosA}]\\=sinAcosA[\frac{1+sinA+cosA+1+sinA-cosA}{(1+sinA)²-cos²A}]\\=sinAcosA[\frac{2(1+sinA)}{1+sin²A+2SinA-cos²A}]\\=sinAcosA[2(\frac{1+SinA)}{1+sin²A+2SinA-1+sin²A}]\\=SinA cosA[\frac{2(1+sinA)}{2sinA(1+sinA)}]\\=cosA$

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