Question

Prove the following identities:

`(i) (sinθ + cosecθ)^2 + (cosθ + secθ)^2 = 7 + tan^2 θ + cot^2 θ`

`(ii) (sinθ + secθ)^2 + (cosθ + cosecθ)^2 = (1 + secθ cosecθ)^2`

`(iii) sec^4 θ– sec^2 θ = tan^4 θ + tan^2 θ`

Answer 1

(i) We have,

LHS = (sinθ + cosecθ)^2 + (cosθ + secθ)^2

`= (sin^2 θ + cosec^2 θ + 2sinθ cosecθ) (cos^2 θ + sec^2 θ + 2cosθ secθ)( \sin ^{2}\theta +\cos ec^{2}\theta +2\sin \theta.\frac{1}{\sin \theta })+( \cos ^{2}\theta +\sec^{2}\theta +2\cos \theta .\frac{1}{\cos \theta })`

`= (sin^2 θ + cosec^2 θ + 2) + (cos^2 θ + sec^2 θ + 2)`

`= sin^2 θ + cos^2 θ + cosec^2 θ + sec^2 θ + 4`

`= 1 + (1 + cot^2 θ) + (1 + tan^2 θ) + 4`

`[∵ cosec2θ = 1 + cot^2 θ, sec^2 θ = 1 + tan^2 θ]`

`= 7 + tan^2 θ + cot^2 θ = RHS.`

(ii) We have,

`LHS = (sinθ + secθ)^2 + (cosθ + cosecθ)^2`

`=( \sin \theta +\frac{1}{\cos \theta } )^{2}+(\cos \theta +\frac{1}{\sin \theta } )^{2}`

`=\sin^{2}\theta +\frac{1}(\cos ^{2}\theta) +\frac{2\sin \theta}{\cos \theta }+\cos ^{2}\theta +\frac{1}(\sin ^{2}\theta}+\frac{2\cos \theta }{\sin \theta } `

`=(\sin^{2}\theta +\cos ^{2}\theta )+( \frac{1}\(cos^{2}\theta) +\frac{1}(\sin ^{2}\theta } )+2( \frac{\sin\theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta })`

`=(\sin ^{2}\theta +\cos ^{2}\theta )+(( \sin^{2}\theta +\cos^{2}\theta)/( \sin ^{2}\theta \cos^{2}\theta } )+(2(\sin ^{2}\theta +\cos^{2}\theta))/{\sin \theta \cos \theta }`

`=1+\frac{1}(\sin^{2}\theta \cos ^{2}\theta }+\frac{2}{\sin\theta \cos \theta } `

`=( 1+\frac{1}{\sin \theta \cos \theta })^{2}`

`= (1 + secθ cosecθ)^2 = RHS`

(iii) We have,

`LHS = sec^4 θ– sec^2 θ`

`= sec^2 θ (sec^2 θ – 1) = (1 + tan^2 θ) (1 + tan^2 θ – 1)`

`[ sec^2 θ = 1 + tan^2 θ]`

`= (1 + tan^2 θ) tan^2 θ`

`= tan^4 θ + tan^2 θ = RHS`