Question

Prove that
tan^3x/1+tan^2x +cot^3x/1+cot^2x= secx.cosecx- 2sinx.cosx

Answer 1

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Solution :-

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tan^3x / (1 + tan^2x ) +

cot^3x / ( 1 + cot^2x )

=tan^3x / sec^2x + cot^3x /cosec^2x

=( sin^3x / cos^3x ) / 1 / cos^2x +

( cos^3x/sin^3x ) / 1 / sin^2x

= (sin^3x / cos^3x ) × ( cos^2x / 1 ) +

( cos^3x / sin^3x ) ×( sin^2 x / 1 )

= sin^3x / cosx + cos^3 x / sinx

= sin^4x + cos^4 x / cosx sinx

=( sin^2x)^2 + (cos^2x)^2 +

2sin^2x cos^2x - 2sin^2x cos^2x / cosx sinx

=( sin^2x + cos^2x )^2-2sin^2x cos^2x /cosx sinx

= (1)^2 - 2sin^2x cos^2x / cosx sinx

= 1 / sinx cosx - 2sin^2x cos^2x/ cosx sinx

= secx cosecx - 2sinx cos x

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Answer 2

$\huge\star\mathfrak\blue{{Answer:-}}$

tan^3x / (1 + tan^2x ) +

cot^3x / ( 1 + cot^2x )

=tan^3x / sec^2x + cot^3x /cosec^2x

=( sin^3x / cos^3x ) / 1 / cos^2x +

( cos^3x/sin^3x ) / 1 / sin^2x

= (sin^3x / cos^3x ) × ( cos^2x / 1 ) +

( cos^3x / sin^3x ) ×( sin^2 x / 1 )

= sin^3x / cosx + cos^3 x / sinx

= sin^4x + cos^4 x / cosx sinx

=( sin^2x)^2 + (cos^2x)^2 +

2sin^2x cos^2x - 2sin^2x cos^2x / cosx sinx

=( sin^2x + cos^2x )^2-2sin^2x cos^2x /cosx sinx

= (1)^2 - 2sin^2x cos^2x / cosx sinx

= 1 / sinx cosx - 2sin^2x cos^2x/ cosx sinx

= secx cosecx - 2sinx cos x