Question

Answer 1

a) \(A=\dfrac{\sin25^0}{\cos65^o}\)

\(\Rightarrow A=\dfrac{\sin\left(90^o-65^0\right)}{\cos65^o}=\dfrac{\cos65^o}{\cos65^o}=1\)

b) \(B=\dfrac{\tan50^o}{\cot40^o}\)

\(\Rightarrow B=\dfrac{\tan\left(90^o-40^o\right)}{\cot40^o}=\dfrac{\cot40^o}{\cot40^o}=1\)

Answer 2

\(a.A=\dfrac{sin25^o}{cos65^o}\\ =\dfrac{sin25^o}{sin\left(90^o-65^o\right)}\\ =\dfrac{sin25^o}{sin25^o}=1\\ b.B=\dfrac{tan50^o}{cot40^o}\\ =\dfrac{tan50^o}{tan\left(90^o-40^o\right)}\\ =\dfrac{tan50^o}{tan50^o}\\ =1\\ c.C=\dfrac{sin70^o}{tan70^o}=cos70^o\)