Question

23. (2) Given, ( m=cos theta+sin theta ) and
[
n=sec theta+operatorname{cosec} theta
]
and
[
frac{m}{n}=frac{cos theta+sin theta}{sec theta+operatorname{cosec} theta}
]
We know that, ( frac{1}{sec theta}=cos theta, frac{1}{operatorname{cosec} theta}=sin theta )
[
begin{aligned}
frac{m}{n} &=frac{cos theta+sin theta}{frac{1}{cos theta}+frac{1}{sin theta}}
&=frac{(cos theta+sin theta)}{(cos theta+sin theta)}(cos theta sin theta)
&=cos theta sin theta
end{aligned}
]

# 23. If cos + sino = m, sece + coseco =n, what is m/n? RRB NTPC (Phase I) 2016 (1) 1 (2) sinocos (3) sec ecosece (4) cotetano

Solution