CBSE Class 10 Sample Paper for 2023 Boards - Maths Standard

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

Last updated at Dec. 13, 2024 by

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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Transcript

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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo