Ncert Solutions For Class 10 Maths Ex 8.4

Ncert Solutions For Class 10 Maths Chapter 8 Ex 8.4

Q1) Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Answer

cosec2Acot2A=1
cosec2A = 1 + cot2A
1sin2A = 1 + cot2A
sin2A = 1/(1+cot2A)
sin A= ±11+cot2A
Now,
sin2A=11+cot2A
1cos2A=11+cot2A
cos2A = 111+cot2A
cos2A = (11+cot2A)(1+cot2A)
1sec2A = (cot2A)(1+cotA)
secA = (1+cotA)(cot2A)

 

secA=±1+cot2AcotA

 

also,
tan A = sinAcosAand cot A = cosAsinA

tan A = 1cotA

 

Q2) Write all the other trigonometric ratios of A in terms of sec A.

Answer

We know that,
sec A = 1cosA
cos A = 1secA
also,
cos2A + sin2A = 1
 sin2A = 1 – cos2A
 sin2A = 1 – (1sec2A)
 sin2A = (sec2A1)sec2A

  sin A=±sec2A1secA

also,
sin A = 1cosecA
cosec A = 1sinA

cosec A=±secAsec2A1
Now,
sec2Atan2A = 1
tan2A = sec2A + 1

tan A=sec2A+1
also,
tan A = 1cotA
cot A = 1tanA

  cot A=±1sec2A+1

 

Q3 Evaluate :


(i) (sin263+sin227)(cos217+cos273)
(ii)  sin25cos65++cos25sin65

 Answer

(i) (sin263+sin227)(cos217+cos273)

 

= [sin2(9027)+sin227][cos2(9073)+cos273]
=(cos227+sin227)(sin227+cos273)
= 11 =1          ( becausesin2A+cos2A=1)

(ii) sin25cos65++cos25sin65
=sin(9025)cos65+cos(9065)sin65

=cos65cos65+sin65sin65

 

= cos65+sin65=1

4) Choose the correct option. Justify your choice.
(i) 9 sec2A – 9 tan2A =
(A) 1                 (B) 9              (C) 8                (D) 0
(ii) (1 + tan Θ + sec Θ) (1 + cot Θ – cosec Θ)
(A) 0                 (B) 1              (C) 2                (D) – 1
(iii) (secA + tanA) (1 – sinA) =
(A) secA           (B) sinA        (C) cosecA      (D) cosA

 

(iv) 1+tan2A1+cot2A=

(A) sec2A

(B) -1

(C) cot2A

(D) tan2A

Answer

(i) (B) is correct.

sec2A– 9 tan2A

= 9 (sec2Atan2A                 )
= 9×1 = 9             ( because  sec2Atan2A = 1)

 

(ii) (C) is correct

(1 + tan θ + sec θ) (1 + cot θ – cosec θ)

= (1 + sin θ/cos θ + 1/cos θ) (1 + cos θ/sin θ – 1/sin θ)

= (cosθ+sin θ+1)/cos θ × (sin θ+cos θ-1)/sin θ

= (cosθ+sin θ)2-12/(cos θ sin θ)

= (cos2θ + sin2θ + 2cos θ sin θ -1)/(cos θ sin θ)

= (1+ 2cos θ sin θ -1)/(cos θ sin θ)

= (2cos θ sin θ)/(cos θ sin θ) = 2

 

(iii) (D) is correct.

(secA + tanA) (1 – sinA)

= (1/cos A + sin A/cos A) (1 – sinA)

= (1+sin A/cos A) (1 – sinA)

= (1 – sin2A)/cos A

= cos2A/cos A = cos A

 

(iv) (D) is correct.

1+tan2A/1+cot2A

= (1+1/cot2A)/1+cot2A

= (cot2A+1/cot2A)×(1/1+cot2A)

= 1/cot2A = tan2A

 

Q5) Prove the following identities, where the angles involved are acute angles for which theexpressions are defined.

(i) (cosec θ – cot θ)= (1-cos θ)/(1+cos θ)

(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A

(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]

(iv) (1 + sec A)/sec A = sin2A/(1-cos A)

[Hint : Simplify LHS and RHS separately]

(v) (cos A–sin A+1)/(cosA+sin A–1) = cosec A + cot A,using the identity cosec2A = 1+cot2A.

(vi)1+sinA1sinA=secA+tanA

(vii) (sin θ – 2sin3θ)/(2cos3θ-cos θ) = tan θ
(viii) (sin A + cosec A)+ (cos A + sec A)2 = 7+tan2A+cot2A
(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
[Hint : Simplify LHS and RHS separately]
(x) (1+tan2A/1+cot2A) = (1-tan A/1-cot A)2 = tan2A

Answer

(i) (cosecΘcotΘ)2 = (1-cos θ)/(1+cos θ)
L.H.S. =  (cosecΘcotΘ)2

=(cosec2Θ+cot2Θ2cosecΘcotΘ)

=(1sin2Θ+cos2Θsin2Θ2cosΘsin2Θ)

= (1 + cos2Θ – 2cos θ)/(1 – cos2Θ)
= (1cosΘ)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+sinA)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)/(cosθ-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) × [(sin3Θcos3Θ)/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 = sin2Θ /(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. = sin2Θ /(1-cos A)
= (1 –cos2Θ)/(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)/(cosA+sin A–1) = cosec A + cot A,using the identity cosec2A = 1+cot2A.
L.H.S. = (cos A–sin A+1)/(cosA+sin A–1)
Dividing Numerator and Denominator by sin A,
= (cos A–sin A+1)/sin A/(cosA+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 cosec2Acot2A = 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.

 

(vi)1+sinA1sinA=secA+tanA

Dividing Numerator and Denominator of L.H.S. by cos A,

= 1cosA+sinAcosA1cosAsinAcosA

 

= secA+tanAsecAtanA

 

= secA+tanAsecAtanAXsecA+tanAsecA+tanA

 

=(secA+tanA)2sec2Atan2A

 

=secA+tanA1

= sec A + tan 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)]/[cosθ(2cos2Θ-1)]
= [sin θ(2cos2Θ -1)]/[cos θ(2cos2Θ -1)]
= tan θ = R.H.S.

 

(viii) (sinA+cosecA)2 + (cosA+secA)2 = 7+tan2A +cot2A
L.H.S. =  (sinA+cosecA)2 + (cosA+secA)2
               = (sin2A + cosec2A + 2 sin A cosec A) + (tcos2A + 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) = (1tanA1cotA)2 =tan2A
L.H.S. = (1+tan2A/1+cot2A)
= (1+tan2A/1+1/tan2A)
= 1+tan2A/[(1+tan2A)/tan2A]
= tan2A

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