Question

Prove that
`(i) sec^(2) theta + "cosec"^(2) theta = sec^(2) theta "cosec"^(2) theta `
`(ii) tan^(2) theta - sin^(2)theta = tan^(2)theta sin^(2)theta `
`(iii) tan^(2) theta + cot^(2)theta +2 = sec^(2) theta "cosec"^(2) theta `

Answer 1

We have
` (i) "LHS "= sec^(2)theta+ "cosec"^(2) theta `
`=(1)/(cos^(2)theta)+ (1)/(sin^(2)theta)=(sin^(2)theta+ cos^(2)theta)/(cos^(2)theta sin^(2)theta ) `
`= (1)/(cos^(2)theta sin^(2)theta) " "[ because sin^(2)theta + cos^(2) theta =1 ] `
` = sec^(2)theta "cosec"^(2)theta = "RHS. " `
`therefore "LHS " = " RHS." `
`(ii) " LHS " = tan^(2)theta- sin^(2)theta `
` =(sin^(2)theta)/(cos^(2)theta)- sin^(2)theta = (sin^(2)theta-sin^(2)theta cos^(2)theta )/(cos^(2)theta) `
` =(sin^(2)theta(1- cos^(2)theta))/(cos^(2)theta)= (sin^(2)theta)/(cos^(2)theta)* sin^(2)theta `
` = tan^(2)theta sin^(2)theta= "RHS." `
` therefore "LHS " = " RHS. "`
` (iii) "LHS "= tan^(2)theta+ cot^(2)theta +2 `
` = (1+tan^(2)theta)+(1+cot^(2)theta)=sec^(2)theta+ "cosec"^(2)theta `
` =(1)/(cos^(2)theta)+(1)/(sin^(2)theta)=(sin^(2)theta+ cos^(2)theta)/(cos^(2)theta sin^(2)theta )`
` =(1)/(cos^(2)theta sin^(2)theta)= sec^(2)theta "cosec"^(2)theta = "RHS. "`
` therefore "LHS " = " RHS. " `