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Ex 8.4, 5  - Prove the following identities (i) cosec - cot - Evaluating using Trignometric Identities

  1. Chapter 8 Class 10 Introduction to Trignometry
  2. Serial order wise
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Ex 8.4, 5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (i) (cosec θ – cot θ)2 = (1 − 𝑐𝑜𝑠" " θ)/(1 + cos⁡θ ) Solving L.H.S (cosec θ – cot θ)2 We need to make it in terms of cos θ & sin θ = (1/sin⁡𝜃 − cos⁡𝜃/sin⁡𝜃 )^2 = (("1 − " cos⁡𝜃)/sin⁡𝜃 )^2 = ((1 − cos⁡〖𝜃)2〗)/(𝑠𝑖𝑛2 𝜃) = ((1 − 〖cos 𝜃〗⁡〖)2〗)/(1 − 𝑐𝑜𝑠2𝜃) = ((1 −〖 cos 𝜃〗⁡〖)2〗)/(12 − 𝑐𝑜𝑠2𝜃) = (1− cos 𝜃)^2/((1 + cos⁡𝜃)(1 − cos⁡𝜃)) = ((1 − 〖cos 𝜃)(1 − 𝑐𝑜𝑠 𝜃)〗⁡ )/((1 + cos⁡𝜃)(1 − cos⁡𝜃)) = (1 −〖 cos〗⁡𝜃)/(1 +〖 cos〗⁡𝜃 ) = RHS Hence proved Ex 8.4, 5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (ii) "cos A" /"1 + sin A" +"1 + sin A" /"cos A" =2 sec A Taking L.H.S (cos⁡ 𝐴)/(1 + sin⁡〖 𝐴〗 )+(1 + sin⁡ 𝐴)/(cos⁡ 𝐴) = (cos⁡ 𝐴 (cos⁡ 𝐴) + (1 + sin⁡ 𝐴)(1 + sin⁡〖 𝐴)〗)/((1 + sin⁡ 𝐴)(cos⁡ 𝐴)) = (𝑐𝑜𝑠2𝐴 + (1 + sin⁡𝐴 )2)/((1 + sin⁡ 𝐴)(cos⁡𝐴)) = (𝑐𝑜𝑠2 𝐴 + 1^2 + 𝑠𝑖𝑛2 𝐴 + 2 sin⁡𝐴)/((1 + sin⁡ 𝐴)(cos⁡𝐴)) = ((𝑐𝑜𝑠2 𝐴 + 𝑠𝑖𝑛2 𝐴) + 1 + 2 sin⁡ 𝐴)/((1 + sin⁡〖 𝐴)(cos⁡ 𝐴)〗 ) = (1 + 1 + 2 sin⁡𝐴)/((1 + sin⁡〖 𝐴)(cos⁡ 𝐴)〗 ) = (2 + 2 sin⁡ 𝐴)/((1 + sin⁡〖 𝐴)(cos⁡ 𝐴)〗 ) = (2(1+ sin⁡ 𝐴))/((1 + sin⁡〖 𝐴)(cos⁡ 𝐴)〗 ) = 2/(cos⁡ 𝐴) = 2 ×1/cos⁡〖 𝐴〗 = 2 sec A = R.H.S ∴ L.H.S = R.H.S Hence proved Ex 8.4, 5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (iii)tan⁡θ/(〖1 − cot〗⁡θ " " )+cot⁡θ/(1 − tan⁡θ ) =1+ sec θ cosec θ [Hint : Write the expression in terms of sin θ and cos θ] Taking L.H.S tan⁡θ/(〖1 − cot〗⁡θ " " )+cot⁡θ/(1 − tan⁡θ ) = (sin⁡θ/cos⁡θ )/(1 − (cos⁡θ/sin⁡θ ) )+((cos⁡θ/sin⁡θ ))/(1 − (sin⁡θ/cos⁡θ ) ) = (sin⁡θ/cos⁡θ )/(((sin⁡θ − cos⁡θ)/sin⁡θ ) )+((cos⁡θ/sin⁡θ ))/( ((cos⁡θ − sin⁡θ)/cos⁡θ ) ) = sin⁡θ/cos⁡θ × sin⁡θ/(sin⁡θ −〖 cos〗⁡θ ) +cos⁡θ/sin⁡θ ×cos⁡θ/(cos⁡θ −〖 sin〗⁡θ ) = sin2⁡θ/(〖cos⁡θ ( sin〗⁡θ −〖 cos〗⁡〖θ )〗 ) + cos2⁡θ/(〖sin⁡θ (cos〗⁡θ −〖 sin〗⁡〖θ )〗 ) = sin2⁡θ/(〖cos⁡θ ( sin〗⁡θ −〖 cos〗⁡〖θ )〗 ) + cos2⁡θ/(〖sin⁡〖θ ×〗 − ( sin〗⁡θ −〖 cos〗⁡〖θ )〗 ) = sin2⁡θ/(〖cos⁡θ (sin〗⁡θ −〖 cos〗⁡〖θ )〗 ) − cos2⁡θ/(〖sin⁡θ ( sin〗⁡θ − cos⁡〖θ )〗 ) = (sin2⁡θ × (sin⁡θ ) −〖 cos2〗⁡θ × (cos⁡θ ))/(〖cos⁡θ ( sin〗⁡θ −〖 cos〗⁡〖θ )sin⁡θ 〗 ) = (sin3⁡θ − cos3⁡θ )/(〖cos⁡θ sin⁡θ ( sin〗⁡θ −〖 cos〗⁡〖θ)〗 ) = (〖(sin〗⁡θ − cos⁡θ)〖(sin2〗⁡θ+ cos2⁡〖θ +cos⁡θ sin⁡θ 〗))/(〖cos⁡θ sin⁡θ ( sin〗⁡θ −〖 cos〗⁡〖θ ) 〗 ) = (〖(sin2〗⁡θ+ cos2⁡〖θ +cos⁡θ sin⁡θ 〗))/(cos⁡θ sin⁡θ ) = (〖(sin2〗⁡θ+ cos2⁡〖θ ) +〖 cos〗⁡θ sin⁡θ 〗)/(cos⁡θ sin⁡θ ) = (1 + cos⁡θ sin⁡θ)/(cos⁡θ sin⁡θ ) = (1 )/(cos⁡θ sin⁡θ ) + (cos⁡θ sin⁡θ)/(cos⁡θ sin⁡θ ) = (1 )/cos⁡θ × (1 )/sin⁡θ + 1 = sec θ × cosec θ + 1 = 1 + sec θ cosec θ = R.H.S ∴ , L.H.S = R.H.S Hence proved Ex 8.4, 5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined. ("1+sec" A)/"sec A" ="sin2 A" /"1 – cos A" "[Hint : Simplify LHS and RHS separately]" Hence L.H.S = R.H.S Hence proved Ex 8.4, 5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined. "cos A – sin A + 1" /"cos A + sin A – 1" = cosec A + cot A, using the identity cosec2 A = 1 + cot2 A. Taking L.H.S (cos⁡𝐴 − sin⁡𝐴 + 1)/(cos⁡𝐴 + sin⁡𝐴 − 1) divide both numerator and denominator by sin A = (1/sin⁡〖 𝐴〗 (cos⁡〖 𝐴 − sin⁡〖𝐴 + 1〗 〗 ))/(1/sin⁡〖 𝐴〗 (cos⁡〖 𝐴 + sin⁡〖 𝐴 − 1〗 〗 ) ) = (cos⁡〖 𝐴〗/sin⁡〖 𝐴〗 − sin⁡〖 𝐴〗/sin⁡〖 𝐴〗 + 1/sin⁡〖 𝐴〗 )/(cos⁡〖 𝐴〗/sin⁡〖 𝐴〗 + sin⁡〖 𝐴〗/sin⁡〖 𝐴〗 − 1/sin⁡〖 𝐴〗 ) = cot⁡〖 𝐴 − 1 + 𝑐𝑜𝑠𝑒𝑐 𝐴〗/cot⁡〖 𝐴 + 1 − 𝑐𝑜𝑠𝑒𝑐 𝐴〗 = ((cot⁡〖 𝐴 + 𝑐𝑜𝑠𝑒𝑐 𝐴) − 1〗)/((cot⁡〖 𝐴 + 1− 𝑐𝑜𝑠𝑒𝑐 𝐴) 〗 ) = ((co𝑡⁡〖 𝐴 + 𝑐𝑜𝑠𝑒𝑐 𝐴) − (𝑐𝑜𝑡2 𝐴 − 𝑐𝑜𝑠𝑒𝑐2 𝐴)〗)/((cot⁡ 𝐴 + 1 − 𝑐𝑜𝑠𝑒𝑐 𝐴)) = ((co𝑡⁡〖 𝐴 + 𝑐𝑜𝑠𝑒𝑐 𝐴) − (cot⁡𝐴 − 𝑐𝑜𝑠𝑒𝑐 𝐴)(cot⁡𝐴 + 𝑐𝑜𝑠𝑒𝑐 𝐴)〗)/((cot⁡ 𝐴 + 1 − 𝑐𝑜𝑠𝑒𝑐 𝐴)) = ((co𝑡⁡〖 𝐴 + 𝑐𝑜𝑠𝑒𝑐 𝐴) [1 − (𝑐𝑜𝑡 𝐴 − 𝑐𝑜𝑠𝑒𝑐 𝐴 )]〗)/([cot⁡ 𝐴 + 1 − 𝑐𝑜𝑠𝑒𝑐 𝐴]) = ((co𝑡⁡〖 𝐴 + 𝑐𝑜𝑠𝑒𝑐 𝐴) [1 − 𝑐𝑜𝑠𝑒𝑐 𝐴 + 𝑐𝑜𝑡 𝐴]〗)/([cot⁡ 𝐴 + 1 − 𝑐𝑜𝑠𝑒𝑐 𝐴]) = cot A + cosec A = R.H.S Hence proved Ex 8.4, 5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (vi) √((1 + sin⁡𝐴 )/(1 −〖 sin〗⁡𝐴 )) = sec A + tan A Taking L.H.S √((1 + sin⁡𝐴 )/(1 −〖 sin〗⁡𝐴 )) Rationalizing denominator Multiplying (1 + sin A) in numerator and denominator = √(((1 + sin⁡𝐴)(1 + sin⁡〖𝐴)〗 )/((1 − sin⁡𝐴)(1 + sin⁡〖𝐴)〗 )) = √(((1 + sin⁡𝐴 )2 )/(12 − 𝑠𝑖𝑛2𝐴)) = √(((1 + sin⁡𝐴 )2 )/(1 − 𝑠𝑖𝑛2𝐴)) =√(((1 +sin⁡𝐴)2 )/(𝑐𝑜𝑠2 𝐴)) =√(((1 + 𝑠𝑖𝑛𝐴 )/(𝑐𝑜𝑠 𝐴))^2 ) = (1 + sin⁡〖 𝐴〗)/cos⁡〖 𝐴〗 = 1/cos⁡〖 𝐴〗 + sin⁡〖 𝐴〗/cos⁡〖 𝐴〗 = sec A + tan A = R.H.S Hence proved Ex 8.4, 5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (vii) (sin θ − 2 sin3 θ)/(2 cos3 θ − cos θ)=tan θ Taking L.H.S (sin θ − 2 sin3θ)/(2 cos3 θ − cosθ) = (sin⁡θ (1 − 2 sin2 θ))/(cos θ(2cos2 θ − 1)) = sin⁡θ/(cos θ) × ( (1 − 2sin2θ))/((2cos2θ − 1)) = tan θ × ( (1 − 2sin2θ))/((2cos2θ − 1)) = tan θ × ((1 − 2sin2θ))/((2(1 − sin2θ) −1) ) = tan θ × ((1 − 2sin2θ))/((2 − 2sin2θ − 1)) = tan θ × ((1 − 2sin2θ))/((1 − 2sin2θ) ) = tan θ × 1 = tan θ = R.H.S Hence proved Ex 8.4, 5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A Taking L.H.S (sin A + cosec A)2 + (cos A + sec A)2 = (sin2 A + cosec2 A + 2sin A cosec A) + (cos2 A + sec2 A + 2 cos A . sec A) = (sin2 A + cosec2 A + 2sin A . 1/sin⁡〖 𝐴〗 ) + (cos2 A + sec2 A + 2 cos A.1/cos⁡〖 𝐴〗 ) = (sin2 A + cosec2 A + 2) + (cos2 A + sec2 A + 2) = (sin2 A + (1 + cot2 A) +2) + (cos2 A +(1 + tan2 A) + 2) = sin2 A + cos2 A + 1 + cot2 A + 2 + 1 + tan2 A + 2 = (sin2 A + cos2 A) + cot2 A + tan2 A + (1 + 2 + 1 + 2) = 1 + cot2 A + tan2 A + 6 = 7 + cot2 A+ tan2 A = R.H.S Hence proved Ex 8.4, 5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (ix) (cosec A – sin A)(sec A – cos A) = 1/(𝑡𝑎𝑛 𝐴 +cot⁡ 𝐴) [Hint : Simplify LHS and RHS separately] Taking L.H.S (cosec A – sin A) (sec A – cos A) = (1/sin⁡〖 𝐴〗 − sin⁡𝐴 )(1/cos⁡〖 𝐴〗 − cos⁡ 𝐴) = ((1 − 𝑠𝑖𝑛2 𝐴))/sin⁡〖 𝐴〗 × ((1 − 𝑐𝑜𝑠2 𝐴))/cos⁡〖 𝐴〗 = 𝑐𝑜𝑠2𝐴/sin⁡〖 𝐴〗 × (𝑠𝑖𝑛2 𝐴)/cos⁡〖 𝐴〗 = sin A cos A Taking R.H.S 1/(𝑡𝑎𝑛 𝐴 +cot⁡ 𝐴) = 1/(sin⁡𝐴/cos⁡𝐴 + cos⁡𝐴/sin⁡𝐴 ) = 1/(sin⁡〖𝐴 (sin⁡〖𝐴) + cos⁡〖𝐴 (cos⁡〖𝐴)〗 〗 〗 〗/cos⁡〖𝐴 sin⁡𝐴 〗 ) = 1/((𝑠𝑖𝑛2 𝐴 + 𝑐𝑜𝑠2 𝐴)/cos⁡〖𝐴 sin⁡𝐴 〗 ) = sin⁡〖 𝐴 . cos⁡ 𝐴〗/(𝑠𝑖𝑛2 𝐴 + 𝑐𝑜𝑠2 𝐴) = sin⁡〖 𝐴 . 〖 cos〗⁡ 𝐴〗/1 = sin A cos A = L.H.S Hence proved Ex 8.4, 5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined. ((1 +𝑡𝑎𝑛2 𝐴)/(1 + 𝑐𝑜𝑡2 𝐴))=((1 −tan⁡〖 𝐴〗)/(1 −cot⁡ 𝐴))^2=𝑡𝑎𝑛2 𝐴 Solving L.H.S ((1 + 𝑡𝑎𝑛2 𝐴)/(1 + 𝑐𝑜𝑡2 𝐴)) = ((1 + 𝑡𝑎𝑛2 𝐴))/(((1+ 1/(𝑡𝑎𝑛2 𝐴)) ) = ((1 + 𝑡𝑎𝑛2 𝐴))/(((𝑡𝑎𝑛2 𝐴 + 1))/(𝑡𝑎𝑛2 𝐴)) = (𝑡𝑎𝑛2 𝐴 (1 + 𝑡𝑎𝑛2 𝐴))/((𝑡𝑎𝑛2 𝐴 + 1)) = tan2 A = R.H.S Now, ((1− tan⁡𝐴)/(1− cot⁡𝐴 ))^2 = ((1 − tan⁡〖 𝐴〗)/(1 − 1/tan⁡〖 𝐴〗 ) " " )^2 = (((1 − tan⁡〖 𝐴)〗)/(((tan⁡〖 𝐴 −1〗 ))/tan⁡〖 𝐴〗 ))^2 = (tan⁡〖 𝐴(1 − tan⁡〖 𝐴)〗 〗/( (tan⁡〖 𝐴 −1)〗 ))^2 = (tan⁡〖 𝐴(1 − tan⁡〖 𝐴)〗 〗/(−(1 − tan⁡〖 𝐴)〗 ))^2 = (−tan⁡𝐴 )^2 = tan2 A

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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