Question

Question 5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

Class 10 - Math - Introduction to Trigonometry Page 193

Answer 1

(i) (cosec θ - cot θ)2 = (1-cos θ)/(1+cos θ)
L.H.S. =  (cosec θ - cot θ)2
           = (cosec2θ + cot2θ - 2cosec θ cot θ)
           = (1/sin2θ + cos2θ/sin2θ - 2cos θ/sin2θ)
           = (1 + cos2θ - 2cos θ)/(1 - cos2θ)
           = (1-cos θ)2/(1 - cosθ)(1+cos θ)
           = (1-cos θ)/(1+cos θ) = R.H.S.

(ii)  cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A
 L.H.S. = cos A/(1+sin A) + (1+sin A)/cos A
            = [cos2A + (1+sin A)2]/(1+sin A)cos A
            = (cos2A + sin2A + 1 + 2sin A)/(1+sin A)cos A
            = (1 + 1 + 2sin A)/(1+sin A)cos A
            = (2+ 2sin A)/(1+sin A)cos A
            = 2(1+sin A)/(1+sin A)cos A
            = 2/cos A = 2 sec A = R.H.S.

(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ
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]
           = sin2θ/[cos θ(sin θ-cos θ)] + cos2θ/[sin θ(cos θ-sin θ)]
           = sin2θ/[cos θ(sin θ-cos θ)] - cos2θ/[sin θ(sin θ-cos θ)]
           = 1/(sin θ-cos θ) [(sin2θ/cos θ) - (cos2θ/sin θ)]
           = 1/(sin θ-cos θ) × [(sin3θ - cos3θ)/sin θ cos θ]
           = [(sin θ-cos θ)(sin2θ+cos2θ+sin θ cos θ)]/[(sin θ-cos θ)sin θ cos θ]
           = (1 + sin θ cos θ)/sin θ cos θ
           = 1/sin θ cos θ + 1
           = 1 + sec θ cosec θ = R.H.S.

(iv)  (1 + sec A)/sec A = sin2A/(1-cos A) 
L.H.S. = (1 + sec A)/sec A
           = (1 + 1/cos A)/1/cos A
           = (cos A + 1)/cos A/1/cos A
           = cos A + 1
R.H.S. = sin2A/(1-cos A)
            = (1 - cos2A)/(1-cos A)
            = (1-cos A)(1+cos A)/(1-cos A)
            = cos A + 1
L.H.S. = R.H.S.

(v) (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A,using the identity cosec2A = 1+cot2A.
L.H.S. = (cos A–sin A+1)/(cos A+sin A–1)
           Dividing Numerator and Denominator by sin A,
           = (cos A–sin A+1)/sin A/(cos A+sin A–1)/sin A
           = (cot A - 1 + cosec A)/(cot A+ 1 – cosec A)
           = (cot A - cosec2A + cot2A + cosec A)/(cot A+ 1 – cosec A) (using cosec2A - cot2A = 1)
           = [(cot A + cosec A) - (cosec2A - cot2A)]/(cot A+ 1 – cosec A)
           = [(cot A + cosec A) - (cosec A + cot A)(cosec A - cot A)]/(1 – cosec A + cot A)
           =  (cot A + cosec A)(1 – cosec A + cot A)/(1 – cosec A + cot A)
           =  cot A + cosec A = R.H.S.


(vii) (sin θ - 2sin3θ)/(2cos3θ-cos θ) = tan θ
L.H.S. = (sin θ - 2sin3θ)/(2cos3θ - cos θ)
           = [sin θ(1 - 2sin2θ)]/[cos θ(2cos2θ- 1)]
           = sin θ[1 - 2(1-cos2θ)]/[cos θ(2cos2θ -1)]
          = [sin θ(2cos2θ -1)]/[cos θ(2cos2θ -1)]
          = tan θ = R.H.S.

(viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7+tan2A+cot2A
L.H.S. = (sin A + cosec A)2 + (cos A + sec A)2
               = (sin2A + cosec2A + 2 sin A cosec A) + (cos2A + sec2A + 2 cos A sec A)
           = (sin2A + cos2A) + 2 sin A(1/sin A) + 2 cos A(1/cos A) + 1 + tan2A + 1 + cot2A
           = 1 + 2 + 2 + 2 + tan2A + cot2A
           = 7+tan2A+cot2A = R.H.S.

(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
L.H.S. = (cosec A – sin A)(sec A – cos A)
           = (1/sin A - sin A)(1/cos A - cos A)
           = [(1-sin2A)/sin A][(1-cos2A)/cos A]
           = (cos2A/sin A)×(sin2A/cos A)
           = cos A sin A
R.H.S. = 1/(tan A+cotA)
            = 1/(sin A/cos A +cos A/sin A)
            = 1/[(sin2A+cos2A)/sin A cos A]
            = cos A sin A
L.H.S. = R.H.S.

(x)  (1+tan2A/1+cot2A) = (1-tan A/1-cot A)2 = tan2A
L.H.S. = (1+tan2A/1+cot2A)
           = (1+tan2A/1+1/tan2A)
           = 1+tan2A/[(1+tan2A)/tan2A]
           = tan2A