NCERT Solutions For Class 10 Math Chapter – 8 Exercise – 8.4

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NCERT Solutions For Class 10 Math Chapter – 8 Exercise – 8.4

 

Q1. Express the trigonometry ratio sin A sec A and tan A in terms of cot A.

Solution:-

 

  • Sin A = 1/cosec A  = 1/√ 1  + cot2A
  • Sec A = √1 + tan2A = √cot2 A + 1 /cot A
  • Tan A = 1/cot A

 

Q2. Write all the other trigonometry ratios of ∟A  in terms of sec A.

Solution:-

 

Sin A = tan A cos A = √sec2 A – 1/ sec A

Tan A = √sec2 A – 1  cot A = 1/tan A = 1/√sec2 A – 1

Cosec A = 1/sin A = sec A/√sec2 A – 1

 

Q3. Evaluate:

  • sin2630 + sin2 270 / cos2 170 + cos2 730 
  • Sin 250cos 650 + cos2 730 

Solution:-

 

  • Sin2630 + sin2 270 / cos2 170 +  cos2 730

= sin2(900 – 270) + sin2270/ cos2 170 + cos2 (900 – 170)

= 1

 

  • sin 250 cos 650 + cos 250 sin 650

= sin 250 cos (900 – 250) + cos 250 sin (900 – 250)

= 1

 

Q4. Choose the correct option Justify your choice.

  • 9 sec2A – 9 tan2 A
  • 1     (B)  9    ©  8     (D) 0

 

  • ( 1 + tanθ + sec θ)(1 + cot θ – cosec θ) =
  • 0      (b)   1 © 2       (d) – 1

 

  • (sec A + tan A)(1 – sin A) =
  • Sec A    (b) Sin A     ( c ) Cosec A    (d) Cos A

 

( iv )1+tan2 A/1 + cot2 A =

  • Sec2 A     (b)  -1     © cot2 A     (d) tan2 A

 

Solution:-

 

  • 9
  • 2
  • Cos A
  • Tan2A

 

Q5. 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)

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

 

(vii) (sin θ – 2sin3θ)/(2cos3θ-cos θ) = tan θ
(viii) (sin A + cosec A)2 + (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

Solution:

(i) (cosec θ – cot θ)2 = (1-cos θ)/(1+cos θ)

To prove this, first take the Left-Hand side (L.H.S) of the given equation, to prove the Right Hand Side (R.H.S)

L.H.S. = (cosec θ – cot θ)2

The above equation is in the form of (a-b)2, and expand it

Since (a-b)2 = a2 + b2 – 2ab

Here a = cosec θ and b = cot θ

= (cosec2θ + cot2θ – 2cosec θ cot θ)

Now, apply the corresponding inverse functions and equivalent ratios to simplify

= (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.

Therefore, (cosec θ – cot θ)2 = (1-cos θ)/(1+cos θ)

Hence proved.

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

Now, take the L.H.S of the given equation.

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

Since cos2A + sin2A = 1, we can write it as

= (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.

L.H.S. = R.H.S.

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

Hence proved.

(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)

We know that tan θ =sin θ/cos θ

cot θ = cos θ/sin θ

Now, substitute it in the given equation, to convert it in a simplified form

= [(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.

Therefore, L.H.S. = R.H.S.

Hence proved

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

First find the simplified form of L.H.S

L.H.S. = (1 + sec A)/sec A

Since secant function is the inverse function of cos function and it is written as

= (1 + 1/cos A)/1/cos A

= (cos A + 1)/cos A/1/cos A

Therefore, (1 + sec A)/sec A = cos A + 1

R.H.S. = sin2A/(1-cos A)

We know that sin2A = (1 – cos2A), we get

= (1 – cos2A)/(1-cos A)

= (1-cos A)(1+cos A)/(1-cos A)

Therefore, sin2A/(1-cos A)= cos A + 1

L.H.S. = R.H.S.

Hence proved

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

With the help of identity function, cosec2A = 1+cot2A, let us prove the above equation.

L.H.S. = (cos A–sin A+1)/(cos A+sin A–1)

Divide the numerator and denominator by sin A, we get

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

We know that cos A/sin A = cot A and 1/sin A = cosec 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.

Therefore, (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A

Hence Proved

 

First divide the numerator and denominator of L.H.S. by cos A,

 

We know that 1/cos A = sec A and sin A/ cos A = tan A and it becomes,

= √(sec A+ tan A)/(sec A-tan A)

Now using rationalization, we get

 

= (sec A + tan A)/1

= sec A + tan A = R.H.S

Hence proved

(vii) (sin θ – 2sin3θ)/(2cos3θ-cos θ) = tan θ

L.H.S. = (sin θ – 2sin3θ)/(2cos3θ – cos θ)

Take sin θ as in numerator and cos θ in denominator as outside, it becomes

= [sin θ(1 – 2sin2θ)]/[cos θ(2cos2θ- 1)]

We know that sin2θ = 1-cos2θ

= sin θ[1 – 2(1-cos2θ)]/[cos θ(2cos2θ -1)]

= [sin θ(2cos2θ -1)]/[cos θ(2cos2θ -1)]

= tan θ = R.H.S.

Hence proved

(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

It is of the form (a+b)2, expand it

(a+b)2 =a2 + b2 +2ab

= (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.

Therefore, (sin A + cosec A)2 + (cos A + sec A)2 = 7+tan2A+cot2A

Hence proved.

(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)

First, find the simplified form of L.H.S

L.H.S. = (cosec A – sin A)(sec A – cos A)

Now, substitute the inverse and equivalent trigonometric ratio forms

= (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

Now, simplify the R.H.S

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.

(cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)

Hence proved

(x)  (1+tan2A/1+cot2A) = (1-tan A/1-cot A)2 = tan2A

L.H.S. = (1+tan2A/1+cot2A)

Since cot function is the inverse of tan function,

= (1+tan2A/1+1/tan2A)

= 1+tan2A/[(1+tan2A)/tan2A]

Now cancel the 1+tan2A terms, we get

= tan2A

(1+tan2A/1+cot2A) = tan2A

Similarly,

(1-tan A/1-cot A)2 = tan2A

Hence proved