Question

prove sin⁴A-cos⁴A+1)cosec²A=2​

Answer 1

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Answer 2

Answer:

$\displaystyle{\boxed{\red{\sf\:(\:\sin^4\:A\:-\:\cos^4\:A\:+\:1\:)\:\times\:cosec^2\:A\:=\:2}}}$

Step-by-step-explanation:

We have given a trigonometric equation.

We have to prove that trigonometric equation.

The given trigonometric equation is

$\displaystyle{\sf\:(\:\sin^4\:A\:-\:\cos^4\:A\:+\:1\:)\:\times\:cosec^2\:A\:=\:2}$

Now,

$\displaystyle{\sf\:(\:\sin^4\:A\:-\:\cos^4\:A\:+\:1\:)\:\times\:cosec^2\:A\:=\:2}$

$\displaystyle{\implies\sf\:LHS\:=\:(\:\sin^4\:A\:-\:\cos^4\:A\:+\:1\:)\:\times\:cosec^2\:A}$

$\displaystyle{\implies\sf\:LHS\:=\:[\:(\:\sin^2\:A\:)^2\:-\:(\:\cos^2\:A\:)^2\:+\:1\:]\:\times\:cosec^2\:A\:\quad\:-\:-\:-\:[\:\because\:(\:a^m\:)^n\:=\:a^{m\:\times\:n}\:]}$

$\displaystyle{\implies\sf\:LHS\:=\:[\:(\:\sin^2\:A\:+\:\cos^2\:A\:)\:(\:\sin^2\:-\:\cos^2\:A\:)\:+\:1\:]\:\times\:cosec^2\:A}$

$\displaystyle{\implies\sf\:LHS\:=\:1\:(\:\sin^2\:A\:-\:\cos^2\:A\:)\:+\:1\:\times\:cosec^2\:A\:\quad\:-\:-\:-\:[\:\because\:\sin^2\:A\:+\:\cos^2\:A\:=\:1\:]}$

$\displaystyle{\implies\sf\:LHS\:=\:sin^2\:A\:-\:\cos^2\:A\:+\:(\:\sin^2\:A\:+\:\cos^2\:A\:)\:\times\:cosec^2\:A\:-\:-\:[\:\sin^2\:A\:+\:\cos^2\:A\:=\:1\:]}$

$\displaystyle{\implies\sf\:LHS\:=\:sin^2\:A\:-\:\cancel{\cos^2\:A}\:+\:\sin^2\:A\:+\:\cancel{\cos^2\:A} </p><p>\:\times\:cosec^2\:A}$

$\displaystyle{\implies\sf\:LHS\:=\:sin^2\:A\:+\:\sin^2\:A\:\times\:cosec^2\:A}$

$\displaystyle{\implies\sf\:LHS\:=\:2\:\cancel{\sin^2\:A}\:\times\:\dfrac{1}{\cancel{\sin^2\:A}}\:\quad\:-\:-\:-\:\left[\:\because\:cosec\:A\:=\:\dfrac{1}{\sin\:A}\:\right]}$

$\displaystyle{\implies\sf\:LHS\:=\:2\:\times\:1}$

$\displaystyle{\implies\sf\:LHS\:=\:2}$

$\displaystyle{\implies\sf\:RHS\:=\:2}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:LHS\:=\:RHS}}}}$

Hence proved!