**Exercise 8.4 Page: 193**

**1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.**

Solution:

To convert the given trigonometric ratios in terms of cot functions, use trigonometric formulas

We know that,

cosec^{2}A– cot^{2}A = 1

cosec^{2}A = 1 + cot^{2}A

Since cosec function is the inverse of sin function, it is written as

1/sin^{2}A = 1 + cot^{2}A

Now, rearrange the terms, it becomes

sin^{2}A = 1/(1+cot^{2}A)

Now, take square roots on both sides, we get

sin A = ±1/(√(1+cot^{2}A)

The above equation defines the sin function in terms of cot function

Now, to express sec function in terms of cot function, use this formula

sin^{2}A = 1/ (1+cot^{2}A)

Now, represent the sin function as cos function

1 – cos^{2}A = 1/ (1+cot^{2}A)

Rearrange the terms,

cos^{2}A = 1 – 1/(1+cot^{2}A)

⇒cos^{2}A = (1-1+cot^{2}A)/(1+cot^{2}A)

Since sec function is the inverse of cos function,

⇒ 1/sec^{2}A = cot^{2}A/(1+cot^{2}A)

Take the reciprocal and square roots on both sides, we get

⇒ sec A = ±√ (1+cot^{2}A)/cotA

Now, to express tan function in terms of cot function

tan A = sin A/cos A and cot A = cos A/sin A

Since cot function is the inverse of tan function, it is rewritten as

tan A = 1/cot A

**2. Write all the other trigonometric ratios of ****∠****A in terms of sec A.**

Solution:

Cos A function in terms of sec A:

sec A = 1/cos A

⇒ cos A = 1/sec A

sec A function in terms of sec A:

cos^{2}A + sin^{2}A = 1

Rearrange the terms

sin^{2}A = 1 – cos^{2}A

sin^{2}A = 1 – (1/sec^{2}A)

sin^{2}A = (sec^{2}A-1)/sec^{2}A

sin A = ± √(sec^{2}A-1)/sec A

cosec A function in terms of sec A:

sin A = 1/cosec A

⇒cosec A = 1/sin A

cosec A = ± sec A/√(sec^{2}A-1)

Now, tan A function in terms of sec A:

sec^{2}A – tan^{2}A = 1

Rearrange the terms

⇒ tan^{2}A = sec^{2}A + 1

tan A = √(sec^{2}A + 1)

cot A function in terms of sec A:

tan A = 1/cot A

⇒ cot A = 1/tan A

cot A = ±1/√(sec^{2}A + 1)

**3. Evaluate:**

**(i) (sin ^{2}63° + sin^{2}27°)/(cos^{2}17° + cos^{2}73°)**

(ii) sin 25° cos 65° + cos 25° sin 65°

Solution:

(i) (sin^{2}63° + sin^{2}27°)/(cos^{2}17° + cos^{2}73°)

To simplify this, convert some of the sin functions into cos functions and cos function into sin function and it becomes,

= [sin^{2}(90°-27°) + sin^{2}27°] / [cos^{2}(90°-73°) + cos^{2}73°)]

= (cos^{2}27°+ sin^{2}27°)/(sin^{2}27° + cos^{2}73°)

= 1/1 =1 (since sin^{2}A + cos^{2}A = 1)

Therefore, (sin^{2}63° + sin^{2}27°)/(cos^{2}17° + cos^{2}73°) = 1

(ii) sin 25° cos 65° + cos 25° sin 65°

To simplify this, convert some of the sin functions into cos functions and cos function into sin function and it becomes,

= sin(90°-25°) cos 65° + cos (90°-65°) sin 65°

= cos 65° cos 65° + sin 65° sin 65°

= cos^{2}65°+ sin^{2}65° = 1 (since sin^{2}A + cos^{2}A = 1)

Therefore, sin 25° cos 65° + cos 25° sin 65° = 1

**4. Choose the correct option. Justify your choice.(i) 9 sec ^{2}A – 9 tan^{2}A =(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) (sec A + tan A) (1 – sin A) =(A) sec A (B) sin A (C) cosec A (D) cos A**

**(iv) 1+tan ^{2}A/1+cot^{2}A = **

** (A) sec ^{2 }A (B) -1 (C) cot^{2}A (D) tan^{2}A**

Solution:

(i) (B) is correct.

Justification:

Take 9 outside, and it becomes

9 sec^{2}A – 9 tan^{2}A

= 9 (sec^{2}A – tan^{2}A)

= 9×1 = 9 (∵ sec2 A – tan2 A = 1)

Therefore, 9 sec^{2}A – 9 tan^{2}A = 9

(ii) (C) is correct

Justification:

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

We know that, tan θ = sin θ/cos θ

sec θ = 1/ cos θ

cot θ = cos θ/sin θ

cosec θ = 1/sin θ

Now, substitute the above values in the given problem, we get

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

Simplify the above equation,

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

= (cos θ+sin θ)^{2}-1^{2}/(cos θ sin θ)

= (cos^{2}θ + sin^{2}θ + 2cos θ sin θ -1)/(cos θ sin θ)

= (1+ 2cos θ sin θ -1)/(cos θ sin θ) (Since cos^{2}θ + sin^{2}θ = 1)

= (2cos θ sin θ)/(cos θ sin θ) = 2

Therefore, (1 + tan θ + sec θ) (1 + cot θ – cosec θ) =2

(iii) (D) is correct.

Justification:

We know that,

Sec A= 1/cos A

Tan A = sin A / cos A

Now, substitute the above values in the given problem, we get

(secA + tanA) (1 – sinA)

= (1/cos A + sin A/cos A) (1 – sinA)

= (1+sin A/cos A) (1 – sinA)

= (1 – sin^{2}A)/cos A

= cos^{2}A/cos A = cos A

Therefore, (secA + tanA) (1 – sinA) = cos A

(iv) (D) is correct.

Justification:

We know that,

tan^{2}A =1/cot^{2}A

Now, substitute this in the given problem, we get

1+tan^{2}A/1+cot^{2}A

= (1+1/cot^{2}A)/1+cot^{2}A

= (cot^{2}A+1/cot^{2}A)×(1/1+cot^{2}A)

= 1/cot^{2}A = tan^{2}A

So, 1+tan^{2}A/1+cot^{2}A = tan^{2}A

**5. Prove the following identities, where the angles involved are acute angles for which theexpressions 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 = sin ^{2}A/(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 cosec ^{2}A = 1+cot^{2}A.**

**(vii) (sin θ – 2sin ^{3}θ)/(2cos^{3}θ-cos θ) = tan θ**

(viii) (sin A + cosec A)^{2 }+ (cos A + sec A)^{2} = 7+tan^{2}A+cot^{2}A

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

[Hint : Simplify LHS and RHS separately]

(x) (1+tan^{2}A/1+cot^{2}A) = (1-tan A/1-cot A)^{2} =tan^{2}A

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} = a^{2} + b^{2} – 2ab

Here a = cosec θ and b = cot θ

= (cosec^{2}θ + cot^{2}θ – 2cosec θ cot θ)

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

= (1/sin^{2}θ + cos^{2}θ/sin^{2}θ – 2cos θ/sin^{2}θ)

= (1 + cos^{2}θ – 2cos θ)/(1 – cos^{2}θ)

= (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)

= [cos^{2}A + (1+sin A)^{2}]/(1+sin A)cos A

= (cos^{2}A + sin^{2}A + 1 + 2sin A)/(1+sin A) cos A

Since cos^{2}A + sin^{2}A = 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]

= sin^{2}θ/[cos θ(sin θ-cos θ)] + cos^{2}θ/[sin θ(cos θ-sin θ)]

= sin^{2}θ/[cos θ(sin θ-cos θ)] – cos^{2}θ/[sin θ(sin θ-cos θ)]

= 1/(sin θ-cos θ) [(sin^{2}θ/cos θ) – (cos^{2}θ/sin θ)]

= 1/(sin θ-cos θ) × [(sin^{3}θ – cos^{3}θ)/sin θ cos θ]

= [(sin θ-cos θ)(sin^{2}θ+cos^{2}θ+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 = sin^{2}A/(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. = sin^{2}A/(1-cos A)

We know that sin^{2}A = (1 – cos^{2}A), we get

= (1 – cos^{2}A)/(1-cos A)

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

Therefore, sin^{2}A/(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 cosec^{2}A = 1+cot^{2}A.

With the help of identity function, cosec^{2}A = 1+cot^{2}A, 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 – cosec^{2}A + cot^{2}A + cosec A)/(cot A+ 1 – cosec A) (using cosec^{2}A – cot^{2}A = 1

= [(cot A + cosec A) – (cosec^{2}A – cot^{2}A)]/(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 – 2sin^{3}θ)/(2cos^{3}θ-cos θ) = tan θ

L.H.S. = (sin θ – 2sin^{3}θ)/(2cos^{3}θ – cos θ)

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

= [sin θ(1 – 2sin^{2}θ)]/[cos θ(2cos^{2}θ- 1)]

We know that sin^{2}θ = 1-cos^{2}θ

= sin θ[1 – 2(1-cos^{2}θ)]/[cos θ(2cos^{2}θ -1)]

= [sin θ(2cos^{2}θ -1)]/[cos θ(2cos^{2}θ -1)]

= tan θ = R.H.S.

Hence proved

(viii) (sin A + cosec A)^{2 }+ (cos A + sec A)^{2} = 7+tan^{2}A+cot^{2}A

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} =a^{2} + b^{2} +2ab

= (sin^{2}A + cosec^{2}A + 2 sin A cosec A) + (cos^{2}A + sec^{2}A + 2 cos A sec A)

= (sin^{2}A + cos^{2}A) + 2 sin A(1/sin A) + 2 cos A(1/cos A) + 1 + tan^{2}A + 1 + cot^{2}A

= 1 + 2 + 2 + 2 + tan^{2}A + cot^{2}A

= 7+tan^{2}A+cot^{2}A = R.H.S.

Therefore, (sin A + cosec A)^{2 }+ (cos A + sec A)^{2} = 7+tan^{2}A+cot^{2}A

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-sin^{2}A)/sin A][(1-cos^{2}A)/cos A]

= (cos^{2}A/sin A)×(sin^{2}A/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/[(sin^{2}A+cos^{2}A)/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+tan^{2}A/1+cot^{2}A) = (1-tan A/1-cot A)^{2} =tan^{2}A

L.H.S. = (1+tan^{2}A/1+cot^{2}A)

Since cot function is the inverse of tan function,

= (1+tan^{2}A/1+1/tan^{2}A)

= 1+tan^{2}A/[(1+tan^{2}A)/tan^{2}A]

Now cancel the 1+tan^{2}A terms, we get

= tan^{2}A

(1+tan^{2}A/1+cot^{2}A) = tan^{2}A

Similarly,

(1-tan A/1-cot A)^{2} =tan^{2}A

Hence proved