Question

ம‌தி‌ப்புகளை‌க் கா‌ண்க


cotθ/tan(90°-θ) + (cot(90°-θ)tanθ•sec(90°-θ))/(sin(90°-θ)cot(90°-θ)•csc(90°-θ) )

Answer 1

tan (90- theeta) = cot theeta

sec (90- theeta) = cosec theeta

sin ( 90- theeta) = cos theeta

opposite is also possible

ANSWER

when we apply all this we get

1+ 1/ cos theeta

Answer 2

cotθ/tan(90°-θ) + (cot(90°-θ)tanθ•sec(90°-θ))/(sin(90°-θ)cot(90°-θ)•csc(90°-θ) ) :

tan($90^o$ -θ ) = cotθ, cos($90^o$ - θ = sinθ)  

sin ($90^o$- θ) = cosθ, cot ($90^o$- θ = tanθ)

cosec ($90^o$ - θ) = secθ, sec ($90^o$ - θ = tan θ)

cosec ($90^o$ - θ) = sec θ, sec ($90^o$- θ = cosec θ)

∴ cot θ / tan($90^o$ - θ) + cot($90^o$ - θ) tan θ sec ($90^o$ θ) / sin ($90^o$ - θ) cot ($90^o$ - θ) cosec ($90^o$ - θ)

= cot θ /cot θ + sin θ tan θ cosec θ / cos θ tan θ sec θ

= 1 + sinθ. 1 / sin θ / cosθ. 1 / cosθ

= 1 + 1

=2  

∴ cosecθ = 1/sinθ secθ

= 1 / cosθ