Question

If m=cosecA sinA and n=secA-tanA prove that (m²n)⅔+(mn²)⅔=1​

Answer 1

sin²A +cos²A = 1

$\red{\boxed{\bf{Step-by-step\: explanation:}}}$

cosecA-sinA=m

→1/sinA sinA=m

→1-sin²A/sinA=m

→cos²A/sinA=m

similarly sin²A/cosA=n

(∛m²n)²=∛m power 4 n²

=∛cos power 8 A * sin power 4 A/sin power 4 A×cos²A

=∛cos power 6 A

=cos²A

similarly (∛mn²)²=sin²A

and (∛m²n)²+(∛mn²) = sin²A+cos²A=1

$\red{\boxed{\tt{Hence\:proved}}}\green\checkmark$

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