Prove that: tan3θ / 1+ tan2θ + cot3θ / 1+ cot2θ = secθ cosecθ

Question 29 Prove the following that- tan^3⁡θ/(1 + 〖𝑡𝑎𝑛^2〗⁡θ ) + cot^3⁡θ/(1 + 〖𝑐𝑜𝑡〗^2⁡θ ) = sec θ cosec θ – 2 sin θ cos θSolving LHS tan^3⁡θ/(1 + 〖𝑡𝑎𝑛^2〗⁡θ ) + cot^3⁡θ/(1 + 〖𝑐𝑜𝑡〗^2⁡θ ) = tan^3⁡θ/(1 + 〖𝑡𝑎𝑛^2〗⁡θ ) + (𝟏/〖𝐭𝐚𝐧〗^𝟑⁡𝜽 )/(𝟏 + 𝟏/〖𝐭𝐚𝐧〗^𝟐⁡𝜽 ) = tan^3⁡θ/(1 + 〖𝑡𝑎𝑛^2〗⁡θ ) + (1/tan^3⁡θ )/((tan^2⁡θ + 1)/tan^2⁡θ ) = tan^3⁡θ/(1 + 〖𝑡𝑎𝑛^2〗⁡θ ) + 1/tan^3⁡θ ×tan^2⁡θ/(tan^2⁡θ + 1) = tan^3⁡θ/(1 + 〖𝑡𝑎𝑛^2〗⁡θ ) + 1/𝑡𝑎𝑛⁡θ ×1/(tan^2⁡θ + 1) = (tan^3⁡θ ×tan⁡〖θ 〗 + 1)/(tan⁡θ (1 + 〖𝑡𝑎𝑛^2〗⁡θ)) = (tan^4⁡θ + 1)/(tan⁡θ (𝟏 + 〖𝒕𝒂𝒏^𝟐〗⁡𝜽)) Putting 1 + tan2 θ = sec2 θ = (tan^4⁡θ + 1)/(𝒕𝒂𝒏⁡𝜽 ×〖𝐬𝐞𝐜〗^𝟐⁡𝜽 ) = (tan^4⁡θ + 1)/(sin⁡θ/cos⁡θ × 1/cos^2⁡θ ) = (tan^4⁡θ + 1)/(sin⁡θ/cos^3⁡θ ) = ((sin⁡𝜃/cos⁡𝜃 )^4 + 1)/(sin⁡θ/cos^3⁡θ ) = (sin^4⁡𝜃/cos^4⁡𝜃 + 1)/(sin⁡θ/cos^3⁡θ ) = ((sin^4⁡𝜃 + cos^4⁡𝜃)/cos^4⁡𝜃 )/(sin⁡θ/cos^3⁡θ ) = (sin^4⁡𝜃 + cos^4⁡𝜃)/cos^4⁡𝜃 ×cos^3⁡θ/sin⁡θ = (〖𝒔𝒊𝒏〗^𝟒⁡𝜽 + 〖𝒄𝒐𝒔〗^𝟒⁡𝜽)/(𝑐𝑜𝑠⁡𝜃 sin⁡θ ) = ((〖𝐬𝐢𝐧〗^𝟐⁡𝜽 + 〖𝐜𝐨𝐬〗^𝟐⁡𝜽 )^𝟐− 𝟐 〖𝐬𝐢𝐧〗^𝟐⁡𝜽 〖𝐜𝐨𝐬〗^𝟐⁡〖𝜽 〗)/(𝑐𝑜𝑠⁡𝜃 sin⁡θ ) Putting 〖𝐬𝐢𝐧〗^𝟐⁡𝜽 + 〖𝐜𝐨𝐬〗^𝟐⁡𝜽 = 1 = (𝟏^𝟐− 2 sin^2⁡𝜃 cos^2⁡〖𝜃 〗)/(𝑐𝑜𝑠⁡𝜃 sin⁡θ ) = (1 − 2 sin^2⁡𝜃 cos^2⁡〖𝜃 〗)/(𝑐𝑜𝑠⁡𝜃 𝑠𝑖𝑛⁡𝜃 ) = 𝟏/(𝒄𝒐𝒔⁡𝜽 𝒔𝒊𝒏⁡𝜽 )−(𝟐 〖𝒔𝒊𝒏〗^𝟐⁡𝜽 〖𝒄𝒐𝒔〗^𝟐⁡〖𝜽 〗)/(𝒄𝒐𝒔⁡𝜽 𝒔𝒊𝒏⁡𝜽 ) = 1/(𝑐𝑜𝑠⁡𝜃 𝑠𝑖𝑛⁡𝜃 )−2 𝑠𝑖𝑛⁡𝜃 𝑐𝑜𝑠⁡〖𝜃 〗 = 𝐬𝐞𝐜 𝜽 𝐜𝐨𝐬𝐞𝐜 𝜽−𝟐 𝒔𝒊𝒏⁡𝜽 𝒄𝒐𝒔⁡〖𝜽 〗 = RHS Hence proved

Question 29 - CBSE Class 10 Sample Paper for 2023 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards

Prove the following that-

tan ³ ⁡θ/1 + tan ² ⁡θ + cot ³ ⁡θ/1 + cot ² ⁡θ = sec θ cosec θ – 2 sin θ cos θ

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