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 `

Prove that `(i) sec^(2) theta + "cosec"^(2) theta = sec^(2) theta "cos

1.	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» 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. " `

`1/(c o s e ctheta-cottheta)-1/(sintheta)=1/(sintheta)-1/(c o s e ctheta+cottheta)`
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