Question

show that √sec^2+cosecA^2=tanA+cotA​

Answer 1

Solution:

$\underline {\bf 1).\;By\;Taking\;LHS:}\\ \\ \\ \implies \sf \sqrt{\sec^{2}A+cosec^{2}\;A} \\ \\ \\ \implies \sf \sqrt{\dfrac{1}{\cos^{2} A}+\dfrac{1}{\sin^{2}A}} \\ \\ \\ \implies \sf \sqrt{\dfrac{\sin^{2}+\cos^{2}A}{\cos^{2}A.\sin^{2}A}}\\ \\ \\ \implies \sf \sqrt{\dfrac{1}{\cos^{2}A.\sin^{2} A}}\\ \\ \\ \implies \sf \dfrac{1}{\cos A.\sin A}\\ \\ \rule{100}{1}\\ \\ \underline {\bf 1).\;By\;Taking\;RHS:}\\ \\ \\ \implies \sf \tan A+\cot A\\ \\ \\ \implies \sf \dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A}$

$\implies \sf \dfrac{\sin^{2}A+\cos^{2}A}{\sin A.\cos A}\\ \\ \\ \implies \sf \dfrac{1}{\sin A.\cos A}\\ \\ \\ \large{\bf LHS=RHS}\\ \\ \\ \huge{\underline{\bf Hence\;Proved}}$