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Page No 288:

Question 1:

If sin θ=32, find the value of all T-ratios of θ.

Answer:

Let us first draw a right ABC, right angled at B and C=θ.
Now, we know that sin θ = perpendicularhypotenuse= ABAC = 32 .

So, if AB = 3k, then AC = 2k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2 
⇒ BC2 = AC2 - AB2 = (2k)2 - (3k)2
⇒ BC2 = 4k2 - 3k2 = k2
⇒ BC = k
Now, finding the other T-ratios using their definitions, we get:
   cos θ  = BCAC = k2k = 12
   tan θ  = ABBC = 3kk = 3

 ∴ cot θ  = 1tan θ = 13, cosec θ = 1sin θ = 23 and sec θ  = 1cos θ = 2

Page No 288:

Question 2:

If cos θ=725  find the values of all T-ratios of θ.

Answer:

Let us first draw a right ABC, right angled at B and C=θ .
Now, we know that cos θ = Basehypotenuse = BCAC  = 725 .

So, if BC = 7k, then AC = 25k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2 
⇒ AB2 = AC2 - BC2 = (25k)2 - (7k)2.
⇒ AB2 = 625k2 - 49k2 = 576k2
⇒ AB = 24k
Now, finding the other trigonometric ratios using their definitions, we get:
   sin θ = ABAC  = 24k25k = 2425 
   tan θ = ABBC = 24k7k = 247 
 ∴ cot θ = 1tan θ = 724 , cosec θ = 1sin θ = 2524  and sec θ  = 1cos θ = 257 

Page No 288:

Question 3:

If tan θ=158 find the values of all T-ratios of θ.

Answer:

Let us first draw a right ABC, right angled at B and C=θ.
Now, we know that tan θ = PerpendicularBase = ABBC = 158.

So, if BC = 8k, then AB = 15k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (15k)2 + (8k)2
⇒ AC2 = 225k2 + 64k2 = 289k2
⇒ AC = 17k

Now, finding the other T-ratios using their definitions, we get:
  sin θ  = ABAC = 15k17k = 1517
  cos θ  = BCAC = 8k17k = 817

∴ cot θ  = 1tan θ = 815, cosec θ = 1sin θ = 1715 and sec θ  = 1cos θ = 178

Page No 288:

Question 4:

If cot θ = 2, find the value of all T-ratios of θ.

Answer:

Let us first draw a right ABC, right angled at B and C=θ.
Now, we know that cot θbasePerpendicular = BCAB = 2.


So, if BC = 2k, then AB = k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (2k)2 + (k)2
⇒ AC2 = 4k2 + k2 = 5k2
⇒ AC = 5k
Now, finding the other T-ratios using their definitions, we get:
   sin θ  = ABAC = k5k = 15
   cos θ  = BCAC = 2k5k = 25

∴ tan θ  = 1cot θ = 12, cosec θ = 1sin θ = 5 and sec θ  = 1cos θ = 52

Page No 288:

Question 5:

If cosec θ = 10, the find the values of all T-ratios of θ.

Answer:

Let us first draw a right ABC, right angled at B and C=θ.
Now, we know that cosec θ = HypotenusePerpendicular = ACAB= 101.

So, if AC = (10)k, then AB = k, where k is a positive number.
Now, by using Pythagoras theorem, we have:
AC2 = AB2 + BC2 
⇒ BC2 = AC2 - AB2 = 10k2 - k2
⇒ BC2 = 9k2
⇒ BC = 3k
Now, finding the other T-ratios using their definitions, we get:
   tan θ  = ABBC = k3k = 13

   cos θ  = BCAC = 3k10k = 310

 ∴ sin θ=1cosec θ=110, cot θ  = 1tan θ = 3 and sec θ  = 1cos θ = 103

Page No 288:

Question 6:

If sinθ=a2-b2a2+b2, find the values of all T-ratios of θ.

Answer:

We have sinθ=a2-b2a2+b2,

As,

cos2θ=1-sin2θ=1-a2-b2a2+b22=11-a2-b22a2+b22=a2+b22-a2-b22a2+b22=a2+b2-a2-b2a2+b2+a2-b2a2+b22
=a2+b2-a2+b2a2+b2+a2-b2a2+b22=2b22a2a2+b22cos2θ=4a2b2a2+b22cosθ=4a2b2a2+b22cosθ=2aba2+b2

Also,

tanθ=sinθcosθ=a2-b2a2+b22aba2+b2=a2-b22ab

Now,

cosecθ=1sinθ=1a2-b2a2+b2=a2+b2a2-b2

Also,

secθ=1cosθ=12aba2+b2=a2+b22ab

And,

cotθ=1tanθ=1a2-b22ab=2aba2-b2

Page No 288:

Question 7:

If 15cotA=8, find the values of sinA and secA.

Answer:

We have,

15cotA=8cotA=815

As,

cosec2A=1+cot2A=1+8152=1+64225=225+64225
cosec2A=289225cosecA=289225cosecA=17151sinA=1715sinA=1517

Also,

cos2A=1-sin2A=1-15172=1-225289=289-225289
cos2A=64289cosA=64289cosA=8171secA=817secA=178

Page No 288:

Question 8:

If sinA=941, find the values of cosA and tanA.

Answer:

We have sinA=941,

As,

cos2A=1-sin2A=1-9412=1-811681=1681-811681
cos2A=16001681cosA=16001681cosA=4041

Also,

tanA=sinAcosA=9414041=940



Page No 289:

Question 9:

If cos θ = 0.6, show that (5 sin θ − 3 tan θ) = 0.

Answer:

Let us consider a right ABC right angled at B.
Now, we know that cos θ = 0.6 = BCAC35

 
 So, if BC = 3k, then AC = 5k, where k is a positive number.
 Using Pythagoras theorem, we have:
AC2 = AB2  + BC2
⇒ AB2 = AC2 - BC2
⇒ AB2 = (5k)2 - (3k)2 = 25k2 - 9k2
⇒ AB2 = 16k2
⇒ AB = 4k

Finding out the other T-ratios using their definitions, we get:
   sin θ = ABAC = 4k5k = 45

   tan θ = ABBC = 4k3k = 43

Substituting the values in the given expression, we get:
 5 sin θ - 3 tan θ
545 - 343 4 - 4 = 0 = RHS
 i.e., LHS  = RHS
 
Hence proved.

Page No 289:

Question 10:

If cosec θ = 2, show that cotθ+sinθ1+cosθ=2.

Answer:

Let us consider a right ABC, right angled at B and C=θ.
Now, it is given that cosec θ = 2.
Also, sin θ  = 1cosecθ = 12 = ABAC


So, if AB = k, then AC = 2k, where k is a positive number.  
Using Pythagoras theorem, we have:
⇒ AC2 = AB2  + BC2
⇒ BC2 = AC2 - AB2
⇒ BC2 = (2k)2 - (k)2
⇒ BC2 = 3k2
⇒ BC = 3k
Finding out the other T-ratios using their definitions, we get:
cos θ = BCAC = 3k2k = 32
tan θ = ABBC = k3k = 13
cot θ = 1tanθ = 3
Substituting these values in the given expression, we get:
cot θ + sinθ1 + cosθ= 3 + 121 + 32= 3 + 122+ 32

=3+12+3=32+3+12+3=23+3+12+3=22+32+3=2
i.e., LHS = RHS

Hence proved.

Page No 289:

Question 11:

If tan θ = 17, show that cosec2θ-sec2θcosec2θ+sec2θ=34.

Answer:

Let us consider a right ABC, right angled at B and C=θ.
Now it is given that tan θABBC17.

So, if AB = k, then BC = 7k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AC2 = (k)2 + (7k)2
⇒ AC2 = k2 + 7k2
⇒ AC = 22k
Now, finding out the values of the other trigonometric ratios, we have:
sin θ  = ABAC = k22k = 122
cos θ  = BCAC = 7 k22k = 722
∴ cosec θ  = 1sin θ = 22 and sec θ   = 1cos θ = 227
Substituting the values of cosec θ  and sec θ  in the given expression, we get:
 cosec2θ - sec2θcosec2θ + sec2θ=(22)2 - 2272(22)2 + 2272=8 - 878 + 87=56 - 8756 + 87=4864 = 34 = RHS
 i.e., LHS = RHS
 
Hence proved.

Page No 289:

Question 12:

If tan θ = 2021, show that1-sinθ+cosθ1+sinθ+cosθ=37.

Answer:

Let us consider a right ABC right angled at B and C=θ.
Now, we know that tan θABBC = 2021

So, if AB = 20k, then BC = 21k, where k is a positive number.
Using Pythagoras theorem, we get:
 AC2 = AB2 + BC2
⇒ AC2= (20k)2 + (21k)2
⇒ AC2 = 841k2
⇒  AC = 29k
Now, sin θ = ABAC = 2029 and cos θ = BCAC = 2129

Substituting these values in the given expression, we get:
  LHS=1 - sinθ + cosθ1 + sinθ + cosθ= 1 - 2029 + 21291 + 2029 + 2129= 29 - 20 + 212929 + 20 + 2129= 3070 = 37 = RHS
∴ LHS = RHS

Hence proved.

Page No 289:

Question 13:

If secθ=54, show that sinθ-2cosθtanθ-cotθ=127.

Answer:

We have,

secθ=541cosθ=54cosθ=45

Also,

sin2θ=1-cos2θ=1-452=1-1625=925sinθ=35

Now,

LHS=sinθ-2cosθtanθ-cotθ=sinθ-2cosθsinθcosθ-cosθsinθ=sinθ-2cosθsin2θ-cos2θsinθ cosθ=sinθ cosθsinθ-2cosθsin2θ-cos2θ
=35×4535-2×45352-452=122535-85925-1625=1225×-55-725=127=RHS

Page No 289:

Question 14:

If cotθ=34, show that secθ-cosecθsecθ+cosecθ=17.

Answer:

LHS=secθ-cosecθsecθ+cosecθ=1cosθ-1sinθ1cosθ+1sinθ=sinθ-cosθsinθ cosθsinθ+cosθsinθ cosθ=sinθ-cosθsinθsinθ+cosθsinθ=sinθsinθ-cosθsinθsinθsinθ+cosθsinθ=1-cotθ1+cotθ=1-341+34
=1474=17=17=RHS

Page No 289:

Question 15:

If sinθ=34, show that cosec2θ-cot2θsec2θ-1=73.

Answer:

LHS=cosec2θ-cot2θsec2θ-1=1tan2θ=cot2θ=cotθ=cosec2θ-1=1sinθ2-1=1342-1=432-1=169-1=16-99=79=73=RHS

Page No 289:

Question 16:

If sinθ=ab, show that secθ+tanθ=b+ab-a.

Answer:

LHS=secθ+tanθ=1cosθ+sinθcosθ=1+sinθcosθ=1+sinθ1-sin2θ=1+ab1-ab2
=11+ab11-a2b2=b+abb2-a2b2=b+abb2-a2b=b+ab+ab-a
=b+ab+ab-a=b+ab-a=b+ab-a=RHS

Page No 289:

Question 17:

If cosθ=35, show that sinθ-cotθ2tanθ=3160.

Answer:

LHS=sinθ-cotθ2tanθ=sinθ-cosθsinθ2sinθcosθ=sin2θ-cosθsinθ2sinθcosθ=cosθsin2θ-cosθ2sin2θ=cosθ1-cos2θ-cosθ21-cos2θ
=351-352-3521-352=3511-925-3521-925=3525-9-1525225-925=3512521625
=35×2×16=3160=RHS

Page No 289:

Question 18:

If tan θ = 43, show that (sin θ + cos θ) = 75.

Answer:

Let us consider a right ABC, right angled at B and C=θ.
Now, we know that  tan θABBC43.

So, if BC = 3k, then AB = 4k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (4k)2 + (3k)2
⇒ AC2 = 16k2 + 9k2 = 25k2
⇒ AC = 5k
Finding out the values of sin θ and cos θ using their definitions, we have:
sin θ = ABAC = 4k5k = 45
cos θ = BCAC = 3k5k = 35
Substituting these values in the given expression, we get:
(sin θ + cos θ) = (45 + 35 ) = (75) = RHS
i.e., LHS = RHS

Hence proved.

Page No 289:

Question 19:

If tan θ = ab, show that a sinθ-b cosθa sinθ+b cosθ=a2-b2a2+b2.

Answer:

It is given that tan θ = ab.

LHS = a sinθ - b cosθa sinθ + b cosθ
 Dividing the numerator and denominator by cos θ, we get:

 a tan θ - ba tan θ + b       (∵ tan θ = sin θcos θ)
Now, substituting the value of tan θ in the above expression, we get:
 aab - baab + b= a2b - ba2b + b= a2 - b2a2 + b2 = RHS
  i.e., LHS = RHS

 Hence proved.

Page No 289:

Question 20:

If 3 tan θ = 4, show that 4cosθ-sinθ2cosθ+sinθ=45.

Answer:

Let us consider a right ABC right angled at B and C=θ.
We know that tan θ = ABBC43

So, if BC = 3k, then AB = 4k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2  + BC2
⇒ AC2 = 16k2 + 9k2
⇒ AC2 = 25k2
⇒ AC = 5k

Now, we have:

sin θ = ABAC = 4k5k = 45

cos θ = BCAC = 3k5k = 35

Substituting these values in the given expression, we get:

 4 cosθ - sinθ2 cosθ + sinθ= 435 - 45235 + 45=125-4565+45= 12 - 456 + 45= 810= 45 = RHS
i.e., LHS = RHS

Hence proved.

Page No 289:

Question 21:

If 3 cot θ = 2, show that 4sinθ-3cosθ2sinθ+6cosθ=13.

Answer:

It is given that cot θ = 23.

LHS  = 4 sinθ - 3 cosθ2 sinθ + 6 cosθ
Dividing the above expression by sin θ, we get:
4 - 3 cot θ2 + 6 cot θ                     [∵ cot θ = cosθsinθ]
Now, substituting the values of cot θ in the above expression, we get:
 4 - 3232 + 623= 4 - 22 + 4 = 26=13
 i.e., LHS = RHS
 
Hence proved.

Page No 289:

Question 22:

If 3cotθ=4, show that 1-tan2θ1+tan2θ=cos2θ-sin2θ.

Answer:

LHS=1-tan2θ1+tan2θ=1-1cot2θ1+1cot2θ=cot2θ-1cot2θcot2θ+1cot2θ=cot2θ-1cot2θ+1=432-1432+1            As, 3cotθ=4 or cotθ=43
=169-1169+1=16-9916+99=79259=725
RHS=cos2θ-sin2θ=cos2θ-sin2θ1=cos2θ-sin2θsin2θ1sin2θ=cos2θsin2θ-sin2θsin2θcosec2θ=cot2θ-1cot2θ+1
=432-1432+1=169-11169+11=16-9916+99=79259=725

Since, LHS=RHS
Hence, verified.

Page No 289:

Question 23:

If sec θ = 178, verify that 3-4sin2θ4cos2θ-3=3-tan2θ1-3tan2θ.

Answer:

It is given that sec θ = 178.

Let us consider a right ABC right angled at B and C=θ.
We know that cos θ = 1sec θ= 817 = BCAC
 
So, if BC = 8k, then AC = 17k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AB2 = AC2 - BC2 = (17k)2 - (8k)2
⇒ AB2 = 289k2 - 64k2 = 225k2
⇒ AB = 15k.

Now, tan θ  = ABBC = 158 and sin θ = ABAC = 15k17k= 1517

The given expression is 3 - 4sin2θ4cos2θ- 3 = 3 - tan2θ1 - 3tan2θ.
 
 Substituting the values in the above expression, we get:
 LHS= 3 - 41517248172 - 3 = 3 - 900289256289- 3 = 867-900256-867= -33-611=33611

RHS = 3-15821-31582=3-225641-67564=192-22564-675=-33-611=33611

∴ LHS = RHS
Hence proved.

Page No 289:

Question 24:

In the adjoining figure, B=90°, BAC=θ°, BC=CD=4 cm and AD=10 cm. Find i sinθ and ii cosθ.

Answer:


In ABD,

Using Pythagoras theorem, we get

AB=AD2-BD2=102-82=100-64=36=6 cm

Again,

In ABC,

Using Pythagoras therem, we get

AC=AB2+BC2=62+42=36+16=52=213 cm

Now,

i sinθ=BCAC=4213=213=21313

ii cosθ=ABAC=6213=313=31313

Page No 289:

Question 25:

In a ∆ABC, it is given that ∠B = 90°, AB = 24 cm and BC = 7 cm.
Figure

Find the value of
(i) sin A
(ii) cos A
(iii) sin C
(iv) cos C

Answer:


Using Pythagoras theorem, we get:
 AC2 = AB2 + BC2
⇒ AC2= (24)2 + (7)2
⇒ AC2 = 576 + 49 = 625
⇒ AC = 25 cm
Now, for T-Ratios of ∠A, base = AB and perpendicular = BC
(i)  sin A = BCAC = 725

(ii) cos A = ABAC = 2425

Similarly, for T-Ratios of ∠C, base = BC and perpendicular = AB
(iii) sin C = ABAC= 2425

(iv) cos C = BCAC = 725



Page No 290:

Question 26:

Given a ∆ABC, in which ∠C = 90°, ∠ABC = θ°, BC = 21 units, AB = 29 units.
Show that (cos2θ − sin2θ) = 41841.

Answer:


Using Pythagoras theorem, we get:
AB2  =  AC2  + BC2
⇒ AC2 = AB2 - BC2
⇒ AC2 = (29)2 - (21)2
⇒ AC2 = 841- 441
⇒ AC2 = 400
⇒ AC = 400 = 20 units

Now, sin θ = ACAB = 2029 and cos θ = BCAB = 2129

 cos2 θ - sin2 θ = 21292 - 20292 =441841-400841=41841

Hence Proved.

Page No 290:

Question 27:

In a ∆ABC, right-angled at B, it is given that AB = 12 cm and BC = 5 cm.

Find the value of
(i) cos A
(ii) cosec A
(iii) cos C
(iv) coses C.

Answer:


Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ AC2 = 122  + 52 = 144 + 25
⇒ AC2 = 169
⇒ AC = 13 cm
Now, for T-Ratios of ∠A, base = AB and perpendicular = BC
(i) cos A = ABAC= 1213
(ii) cosec A = 1sin A = ACBC = 135

Similarly, for T-Ratios of ∠C, base = BC and perpendicular = AB
(iii) cos C = BCAC = 513
(iv) cosec C = 1sin C = ACAB =1312

Page No 290:

Question 28:

If sinα=12, prove that 3cosα-4cos3α=0.

Answer:

LHS=3cosα-4cos3α=cosα3-4cos2α=1-sin2α3-41-sin2α=1-1223-41-122=11-143-411-14=343-434=343-3=340=0=RHS

Page No 290:

Question 29:

In a ABC, B = 90° and tanA = 13. Prove that
(i) sinA·cosC + cosA·sinC = 1            (ii) cosA·cosC - sinA·sinC = 0

Answer:



In ABC, B=90°,As, tanA=13BCAB=13Let BC=x and AB=x3Using Pythagoras theorem, we getAC=AB2+BC2=x32+x2=3x2+x2=4x2=2x

Now,

i LHS=sinA·cosC+cosA·sinC=BCAC·BCAC+ABAC·ABAC=BCAC2+ABAC2=x2x2+x32x2=14+34=44=1=RHS

ii LHS=cosA·cosC-sinA·sinC=ABAC·BCAC-BCAC·ABAC=x32x.x2x-x2x.x32x=34-34=0=RHS

Page No 290:

Question 30:

If A and B are acute angles such that sinA = sinB, then prove that A = B.

Answer:


In ABC, C = 90°
sinA = BCAB and
sinB = ACAB

As, sinA = sinB
BCAB = ACAB
BC = AC
So, A = B             (Angles opposite to equal sides are equal)

Page No 290:

Question 31:

If A and B are acute angles such that tanA = tanB, the prove that A=B.

Answer:



In ABC, C=90°,

tanA=BCAC andtanB=ACBC

As, tanA=tanB
BCAC=ACBCBC2=AC2BC=ACSo, A=B             Angles opposite to equal sides are equal

Page No 290:

Question 32:

In a right ABC, right-angled at B, if tanA=1, then verify that 2sinA·cosA=1.

Answer:

We have,tanA=1sinAcosA=1sinA=cosAsinA-cosA=0Squaring both sides, we getsinA-cosA2=0sin2A+cos2A-2sinA·cosA=01-2sinA·cosA=02sinA·cosA=1

Page No 290:

Question 33:

In the figure of PQR, P=θ° and R=ϕ°. Find
i x+1cotϕii x3+x2tanθiii cosθ

Answer:


In PQR, Q=90°,

Using Pythagoras theorem, we get

PQ=PR2-QR2=x+22-x2=x2+4x+4-x2=4x+1=2x+1

Now,

i x+1cotϕ=x+1×QRPQ=x+1×x2x+1=x2

ii x3+x2tanθ=x2x+1×QRPQ=xx+1×x2x+1=x22

iii cosθ=PQPR=2x+1x+2

Page No 290:

Question 34:

If x=cosecA+cosA and y=cosecA-cosA, then prove that 2x+y2+x-y22-1=0.

Answer:

LHS=2x+y2+x-y22-1=2cosecA+cosA+cosecA-cosA2+cosecA+cosA-cosecA-cosA22-1=2cosecA+cosA+cosecA-cosA2+cosecA+cosA-cosecA+cosA22-1=22cosecA2+2cosA22-1
=1cosecA2+cosA2-1=sinA2+cosA2-1=sin2A+cos2A-1=1-1=0=RHS

Page No 290:

Question 35:

If x=cotA+cosA and y=cotA-cosA, prove that x-yx+y2+x-y22=1.

Answer:

LHS=x-yx+y2+x-y22=cotA+cosA-cotA-cosAcotA+cosA+cotA-cosA2+cotA+cosA-cotA-cosA22=cotA+cosA-cotA+cosAcotA+cosA+cotA-cosA2+cotA+cosA-cotA+cosA22=2cosA2cotA2+2cosA22=cosAcosAsinA2+cosA2=sinA cosAcosA2+cosA2=sinA2+cosA2=sin2A+cos2A=1=RHS



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