Last updated at June 23, 2017 by Teachoo

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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 + cosθ ) Solving L.H.S (cosec θ – cot θ)2 We need to make it in terms of cos θ & sin θ = (1/sin𝜃 − cos𝜃/sin𝜃 )^2 = (("1 − " cos𝜃)/sin𝜃 )^2 = ((1 − cos〖𝜃)2〗)/(𝑠𝑖𝑛2 𝜃) = ((1 − 〖cos 𝜃〗〖)2〗)/(1 − 𝑐𝑜𝑠2𝜃) = ((1 −〖 cos 𝜃〗〖)2〗)/(12 − 𝑐𝑜𝑠2𝜃) = (1− cos 𝜃)^2/((1 + cos𝜃)(1 − cos𝜃)) = ((1 − 〖cos 𝜃)(1 − 𝑐𝑜𝑠 𝜃)〗 )/((1 + cos𝜃)(1 − cos𝜃)) = (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 + sin𝐴 )2)/((1 + sin 𝐴)(cos𝐴)) = (𝑐𝑜𝑠2 𝐴 + 1^2 + 𝑠𝑖𝑛2 𝐴 + 2 sin𝐴)/((1 + sin 𝐴)(cos𝐴)) = ((𝑐𝑜𝑠2 𝐴 + 𝑠𝑖𝑛2 𝐴) + 1 + 2 sin 𝐴)/((1 + sin〖 𝐴)(cos 𝐴)〗 ) = (1 + 1 + 2 sin𝐴)/((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)tanθ/(〖1 − cot〗θ " " )+cotθ/(1 − tanθ ) =1+ sec θ cosec θ [Hint : Write the expression in terms of sin θ and cos θ] Taking L.H.S tanθ/(〖1 − cot〗θ " " )+cotθ/(1 − tanθ ) = (sinθ/cosθ )/(1 − (cosθ/sinθ ) )+((cosθ/sinθ ))/(1 − (sinθ/cosθ ) ) = (sinθ/cosθ )/(((sinθ − cosθ)/sinθ ) )+((cosθ/sinθ ))/( ((cosθ − sinθ)/cosθ ) ) = sinθ/cosθ × sinθ/(sinθ −〖 cos〗θ ) +cosθ/sinθ ×cosθ/(cosθ −〖 sin〗θ ) = sin2θ/(〖cosθ ( sin〗θ −〖 cos〗〖θ )〗 ) + cos2θ/(〖sinθ (cos〗θ −〖 sin〗〖θ )〗 ) = sin2θ/(〖cosθ ( sin〗θ −〖 cos〗〖θ )〗 ) + cos2θ/(〖sin〖θ ×〗 − ( sin〗θ −〖 cos〗〖θ )〗 ) = sin2θ/(〖cosθ (sin〗θ −〖 cos〗〖θ )〗 ) − cos2θ/(〖sinθ ( sin〗θ − cos〖θ )〗 ) = (sin2θ × (sinθ ) −〖 cos2〗θ × (cosθ ))/(〖cosθ ( sin〗θ −〖 cos〗〖θ )sinθ 〗 ) = (sin3θ − cos3θ )/(〖cosθ sinθ ( sin〗θ −〖 cos〗〖θ)〗 ) = (〖(sin〗θ − cosθ)〖(sin2〗θ+ cos2〖θ +cosθ sinθ 〗))/(〖cosθ sinθ ( sin〗θ −〖 cos〗〖θ ) 〗 ) = (〖(sin2〗θ+ cos2〖θ +cosθ sinθ 〗))/(cosθ sinθ ) = (〖(sin2〗θ+ cos2〖θ ) +〖 cos〗θ sinθ 〗)/(cosθ sinθ ) = (1 + cosθ sinθ)/(cosθ sinθ ) = (1 )/(cosθ sinθ ) + (cosθ sinθ)/(cosθ sinθ ) = (1 )/cosθ × (1 )/sinθ + 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 (cos𝐴 − sin𝐴 + 1)/(cos𝐴 + sin𝐴 − 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) − (cot𝐴 − 𝑐𝑜𝑠𝑒𝑐 𝐴)(cot𝐴 + 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 + sin𝐴 )/(1 −〖 sin〗𝐴 )) = sec A + tan A Taking L.H.S √((1 + sin𝐴 )/(1 −〖 sin〗𝐴 )) Rationalizing denominator Multiplying (1 + sin A) in numerator and denominator = √(((1 + sin𝐴)(1 + sin〖𝐴)〗 )/((1 − sin𝐴)(1 + sin〖𝐴)〗 )) = √(((1 + sin𝐴 )2 )/(12 − 𝑠𝑖𝑛2𝐴)) = √(((1 + sin𝐴 )2 )/(1 − 𝑠𝑖𝑛2𝐴)) =√(((1 +sin𝐴)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) = (sinθ (1 − 2 sin2 θ))/(cos θ(2cos2 θ − 1)) = sinθ/(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〗 − sin𝐴 )(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/(sin𝐴/cos𝐴 + cos𝐴/sin𝐴 ) = 1/(sin〖𝐴 (sin〖𝐴) + cos〖𝐴 (cos〖𝐴)〗 〗 〗 〗/cos〖𝐴 sin𝐴 〗 ) = 1/((𝑠𝑖𝑛2 𝐴 + 𝑐𝑜𝑠2 𝐴)/cos〖𝐴 sin𝐴 〗 ) = 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− tan𝐴)/(1− cot𝐴 ))^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 = (−tan𝐴 )^2 = tan2 A

Example 2
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Example 6 Important

Example 7 Important

Example 11 Important

Example 15 Important

Ex 8.1, 5 Important

Ex 8.1, 10 Important

Ex 8.1, 9 Important

Ex 8.3, 5 Important

Ex 8.1, 11 Important

Ex 8.3, 6 Important

Ex 8.3, 7 Important

Ex 8.4, 1 Important

Ex 8.4, 2 Important

Ex 8.4, 5 Important You are here

Important Questions for Exam - Class 10

- Chapter 1 Class 10 Real Numbers
- Chapter 2 Class 10 Polynomials
- Chapter 3 Class 10 Pair of Linear Equations in Two Variables
- Chapter 4 Class 10 Quadratic Equations
- Chapter 5 Class 10 Arithmetic Progressions
- Chapter 6 Class 10 Triangles
- Chapter 7 Class 10 Coordinate Geometry
- Chapter 8 Class 10 Introduction to Trignometry
- Chapter 9 Class 10 Some Applications of Trignometry
- Chapter 10 Class 10 Circles
- Chapter 11 Class 10 Constructions
- Chapter 12 Class 10 Areas related to Circles
- Chapter 13 Class 10 Surface Areas and Volumes
- Chapter 14 Class 10 Statistics
- Chapter 15 Class 10 Probability

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Davneet Singh

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