Question

Without using trigonometric tables, evaluate the following:

`(\sin ^{2}20^\text{o}+\sin^{2}70^\text{o})/(\cos ^{2}20^\text{o}+\cos ^{2}70^\text{o}}+\frac{\sin (90^\text{o}-\theta )\sin \theta }{\tan \theta }+\frac{\cos (90^\text{o}-\theta )\cos \theta }{\cot \theta }`

Answer 1

`(sin^{2}20^\text{o}+\sin ^{2}70^\text{o})/(\cos ^{2}20^\text{o}+\cos ^{2}70^\text{o}}+\frac{\sin (90^\text{o}-\theta )\sin \theta }{\tan \theta }+\frac{\cos (90^\text{o}-\theta )\cos \theta }{\cot \theta }`

`=(\sin^{2}20^\text{o}+\sin^{2}(90^\text{o}-20^\text{o}))/(\cos ^{2}20^\text{o}+\cos^{2}(90^\text{o}-20^\text{o})}+\frac{\sin (90^\text{o}-\theta)\sin \theta }{\tan \theta }+\frac{\cos (90^\text{o}-\theta )\cos\theta }{\cot \theta }`

`=(sin ^{2}20^\text{o}+\cos ^{2}20^\text{o})/(\cos^{2}20^\text{o}+\sin^{2}20^\text{o}}+\frac{\cos \theta \sin\theta }{\frac{\sin \theta }{\cos \theta }}+\frac{\sin \theta \cos\theta }{\frac{\cos \theta }{\sin \theta }}`

`[ \sin (90^\text{o}-\theta )=\cos \theta " and "\cos (90^\text{o}-\theta )=\sin \theta]`

`= 1 + cos^2 θ + sin^2 θ = 1 + 1 = 2`