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**Exercise 8.1 Page No: 89**

**Choose the correct answer from the given four options:****1. If cos A = 4/5, then the value of tan A is****(a) 3/5 ****(b) 3/4****(c) 4/3 ****(d) 5/3****Ans: **(b)**Explanation:**

According to the question,

cos A = 4/5 …(1)

We know,

tan A = sinA/cosA

To find the value of sin A,

We have the equation,

sin^{2} θ +cos^{2} θ = 1

So, sin θ = √ (1-cos^{2} θ)

Then,

sin A = √ (1-cos^{2} A) …(2)

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

sin A = √(1-cos^{2} A)

Substituting equation (1) in (2),

We get,

Sin A = √(1-(4/5)^{2})

= √(1-(16/25))

= √(9/25)

= ¾

Therefore,**2. If sin A = 1/2, then the value of cot A is(a) √3(b) 1/√3(c) √3/2(d) 1**

(A) √3

According to the question,

Sin A = ½ … (1)

We know that,

To find the value of cos A.

We have the equation,

sin

So, cos θ = √(1-sin

Then,

cos A = √(1-sin

cos

cos A = √ (1 - sin

Substituting equation 1 in 3, we get,

cos A = √(1 - 1/4) = √(3/4) = √3/2

Substituting values of sin A and cos A in equation 2, we get

cot A = (√3/2) × 2 = √3

According to the question,

We have to find the value of the equation,

cosec(75° + θ) – sec(15° - θ) – tan(55° + θ) + cot(35° - θ)

= cosec[90°-(15°-θ)] – sec(15°-θ) – tan(55°+θ) + cot[90°-(55°+θ)]

Since, cosec (90°- θ) = sec θ

And, cot(90°-θ) = tan θ

We get,

= sec(15° - θ) – sec(15° - θ) – tan(55° + θ) + tan(55° + θ)

= 0

According to the question,

sin θ = a/b

We know, sin

sin

sin A = √(1-cos

So, cos θ = √(1 - a

Hence, cos θ = √(b

According to the question,

cos(α+β) = 0

Since, cos 90° = 0

We can write,

cos(α+β)= cos 90°

By comparing cosine equation on L.H.S and R.H.S,

We get,

(α + β) = 90°

α = 90°- β

Now we need to reduce sin (α - β),

So, we take,

sin(α - β) = sin(90°- β - β) = sin(90°- 2β)

sin(90° - θ) = cos θ

So, sin(90° - 2β) = cos 2β

Therefore, sin(α - β) = cos 2β

tan 1°. tan 2°.tan 3° …… tan 89°

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.tan 45°.tan 46°.tan 47°…tan 87°.tan 88°.tan 89°

Since, tan 45° = 1,

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.tan 46°.tan 47°…tan 87°.tan 88°.tan 89°

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.tan(90°-44°).tan(90°-43°)…tan(90°-3°). tan(90°-2°).tan(90°-1°)

Since, tan(90°-θ) = cot θ,

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.cot 44°.cot 43°…cot 3°.cot 2°.cot 1°

Since, tan θ = (1/cot θ)

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1. (1/tan 44

= 1

Hence, tan 1°.tan 2°.tan 3° …… tan 89° = 1

According to the question,

cos 9∝ = sin ∝ and 9∝<90°

i.e. 9α is an acute angle

We know that,

sin(90°-θ) = cos θ

So,

cos 9∝ = sin (90°- ∝)

Since, cos 9∝ = sin(90°- 9∝) and sin(90°- ∝) = sin∝

Thus, sin (90°- 9∝) = sin∝

90°- 9∝ = ∝

10∝ = 90°

∝ = 9°

Substituting ∝ = 9° in tan 5∝, we get,

tan 5∝ = tan (5 × 9) = tan 45° = 1

∴, tan 5∝ = 1

**Exercise 8.2 Page No: 93**

**Write ‘True’ or ‘False’ and justify your answer in each of the following:****1. tan 47 ^{o}/cot 43° = 1**

True

Justification:

Since, tan (90° - θ) = cot θ

False

Justification:

Since, (a

cos

= [cos 23°+sin(90°-23°)] [cos 23°-sin(90°-23°)]

= (cos 23°+cos 23°)(cos 23°-cos 23°) (∵sin(90°-θ) = cos θ)

= (cos 23°+cos 23°).0

= 0, which is neither positive nor negative

False

Justification:

We know that,

sin θ increases when 0° ≤ θ ≤ 90°

cos θ decreases when 0° ≤ θ ≤ 90°

And (sin 80°-cos 80°) = (increasing value-decreasing value)

= a positive value.

Therefore, (sin 80°-cos 80°) > 0.

**4. √((1– cos ^{2}θ) sec^{2} θ)= tan θ**

True

Justification:

True

Justification:

According to the question,

cos A+cos

i.e., cos A = 1- cos

Since,

sin

sin

We get,

cos A = sin

Squaring L.H.S and R.H.S,

cos

To find sin

Adding equations (1) and (2),

We get

sin

Therefore, sin

False

Justification:

L.H.S = (tan θ + 2) (2 tan θ + 1)

= 2 tan

= 2 tan

Since, sec

= 2(sec

= 2 sec

= 5 tan θ + 2 sec

∴, L.H.S ≠ R.H.S

**Exercise 8.3 Page No: 95**

**Prove the following (from Q.1 to Q.7):****1. sin θ/(1 + cos θ) + (1 + cos θ)/sin θ = 2cosec θ****Solution:**

L.H.S =

R.H.S

Hence proved.**2. tan A/(1 + secA) – tan A/(1 - secA) = 2cosec A Solution:**

L.H.S:

Since,

sec

sec

= R.H.S

Hence proved.

According to the question,

tan A = ¾

We know,

tan A = perpendicular/ base

So,

tan A = 3k/4k

Where,

Perpendicular = 3k

Base = 4k

Using Pythagoras Theorem,

(hypotenuse)

(hypotenuse)

hypotenuse = 5k

To find sin A and cos A,

Hence, proved.

Solution:

L.H.S:

(sin α + cos α) (tan α + cot α)

As we know,

= R.H.S

Hence, proved.

L.H.S: (√3 + 1) (3 – cot 30°)

= (√3 + 1) (3 – √3) [∵cos 30° = √3]

= (√3 + 1) √3 (√3 – 1) [∵(3 – √3) = √3 (√3 – 1)]

= ((√3)

= (3-1) √3

= 2√3

Similarly solving R.H.S: tan

Since, tan 60

We get,

(√3)

= 2√3

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

Hence, proved.

And, we know that,

L.H.S =

Since, tan (90° – θ) = cot θ

tan θ + tan (90° – θ)= tan θ + cot θ

**Exercise 8.4 Page No: 99**

1. If cosecθ + cotθ = p, then prove that cosθ = (p^{2} – 1)/ (p^{2} + 1).

Solution:

According to the question,

cosec θ + cot θ = p

Since,

Hence, proved.**2. Prove that √(sec ^{2} θ + cosec^{2} θ) = tan θ + cot θ**

L.H.S =

√ (sec

Since,

= R.H.S

Hence, proved.

Let PR = h meter, be the height of the tower.

The observer is standing at point Q such that, the distance between the observer and tower is QR = (20 + x) m, where

QR = QS + SR = 20 + x

∠PQR = 30°

∠ PSR = θ

In ∆PQR,

Rearranging the terms,

We get 20 + x = √3h

⇒ x = √3h – 20 …eq.1

In ∆PSR,

tan θ = h/x

Since, angle of elevation increases by 15

We have,

θ = 30° + 15° = 45°

So,

tan 45

⇒ 1 = h/x

⇒ h = x

Substituting x=h in eq. 1, we get

h = √3 h – 20

⇒ √3 h – h = 20

⇒ h (√3 – 1) = 20

= 10 (√3 + 1)

Hence, the required height of the tower is 10 (√3 + 1) meter.

Given: 1+sin

Dividing L.H.S and R.H.S equations with sin

We get,

cosec

Since,

cosec

⇒ cot

⇒ cot

⇒ cot

Splitting the middle term and then solving the equation,

⇒ cot

⇒ cot θ(cot θ -1)–2(cot θ +1) = 0

⇒ (cot θ – 1)(cot θ – 2) = 0

⇒ cot θ = 1, 2

Since,

tan θ = 1/cot θ

tan θ = 1, ½

Hence, proved.

Given: sin θ + 2 cos θ = 1

Squaring on both sides,

(sin θ +2 cos θ)

⇒ sin

Since, sin

⇒ (1 – cos

⇒ 1 – cos

⇒ – 4 sin

⇒ 4 sin

We know that,

a

So, we get,

(2sin θ – cos θ)

⇒ 2sin θ – cos θ = 2

Hence proved.

Let BC = s; PC = t

Let height of the tower be AB = h.

∠ABC = θ and ∠APC = 90° – θ

(∵ the angle of elevation of the top of the tower from two points P and B are complementary)

⇒ h^{2} = st

⇒ h = √st

Hence the height of the tower is √st.**7. The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation is 30° than when it is 60°. Find the height of the tower.****Solution:****Let SQ = h be the tower. ****∠SPQ = 30° and ∠SRQ = 60° ****According to the question, the length of shadow is 50 m long hen angle of elevation of the sun is 30° than when it was 60°. So, **

PR = 50 m and RQ = x m

⇒ 50√3+h = 3h

⇒ 50√3 = 3h – h

⇒ 3h – h = 50√3

⇒ 2h = 50√3

⇒ h = (50√3)/2

⇒ h = 25√3

Hence, the required height is 25√3 m.**8. A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height h. At a point on the plane, the angles of elevation of the bottom and the top of the flag staff are α and β, respectively. Prove that the height of the tower is [h tan α/(tan β – tan α)].Solution:**

Given that a vertical flag staff of height h is surmounted on a vertical tower of height H(say),

such that FP = h and FO = H.

The angle of elevation of the bottom and top of the flag staff on the plane is ∠PRO = α and ∠FRO = β respectively

In ∆PRO, we have

Hence, proved.**9. If tanθ + secθ = l, then prove that secθ = (l ^{2} + 1)/2l.**

Given: tan θ+ sec θ = l …eq. 1

Multiplying and dividing by (sec θ – tan θ) on numerator and denominator of L.H.S,

So, sec θ – tan θ = 1 …eq.2

Adding eq. 1and eq. 2, we get

(tan θ + sec θ) + (sec θ – tan θ) = 1

Hence, proved.

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