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Page No 546:

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 546:

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 546:

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 546:

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 546:

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 546:

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 546:

Question 7:

If sinθ=cc2+d2, where d > 0 then find the values of cos θ and tan θ.

Answer:

Given: sinθ=cc2+d2Since, sinθ=PHP=c and H=c2+d2Using Pythagoras theorem,P2+B2=H2c2+B2=c2+d2B2=d2B=dTherefore,cosθ=BH=dc2+d2tanθ=PB=cdHence, cosθ=dc2+d2 and tanθ=cd.

Page No 546:

Question 8:

If 3 tan θ=1 then evaluate (cos2θ – sin2θ).

Answer:

Given: 3tanθ=1tanθ=13Since, tanθ=PBP=1 and B=3Using Pythagoras theorem,P2+B2=H212+32=H2H2=1+3=4H=2Therefore,sinθ=PH=12cosθ=BH=32cos2θ-sin2θ=322-122                   =34-14=24                   =12Hence, cos2θ-sin2θ=12.

Page No 546:

Question 9:

If 4tan θ = 3 then prove that sin θ cos θ=1225.

Answer:

Given: 4tanθ=3tanθ=34Since, tanθ=PBP=3 and B=4Using Pythagoras theorem,P2+B2=H232+42=H2H2=9+16=25H=5Therefore,sinθ=PH=35cosθ=BH=45sinθ×cosθ=35×45                   =1225Hence, sinθ×cosθ=1225.

Page No 546:

Question 10:

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 547:

Question 11:

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 547:

Question 12:

If sin θ=1213 then evaluate 2sin θ-3cos θ4sin θ-9cos θ.

Answer:

Given: sinθ=1213Since, sinθ=PHP=12 and H=13Using Pythagoras theorem,P2+B2=H2122+B2=132B2=169-144B2=25B=5Therefore,cosθ=BH=513Now,2sinθ-3cosθ4sinθ-9cosθ=21213-351341213-9513                        =2413-15134813-4513                        =24-1548-45                        =93                        =3Hence, 2sinθ-3cosθ4sinθ-9cosθ=3.

Page No 547:

Question 13:

If tan θ=12 then evaluate cos θsin θ+sin θ1+cos θ

Answer:

Given: tanθ=12Since, tanθ=PBP=1 and B=2Using Pythagoras theorem,P2+B2=H212+22=H2H2=1+4H2=5H=5Therefore,sinθ=PH=15cosθ=BH=25Now,cosθsinθ+sinθ1+cosθ=2515+151+25                             =21+155+25                             =21+15+2                             =2+15+2×5-25-2                             =2+5-25-4                             =2+5-2                             =5Hence, cosθsinθ+sinθ1+cosθ=5.

Page No 547:

Question 14:

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 547:

Question 15:

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 547:

Question 16:

If sec θ = 178 then prove 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 547:

Question 17:

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 547:

Question 18:

If tan θ = 17 then prove that cosec2θ+sec2θcosec2θ-sec2θ=43.

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 547:

Question 19:

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 547:

Question 20:

If 3 tan A = 4 then prove that
(i) sec A-cosec Asec A+cosec A=17
(ii) 1-sin A1+cos A=122

Answer:

(i)
 ​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

(ii)
Given: 3tanA=4tanA=43Since, tanA=PBP=4 and B=3Using Pythagoras theorem,P2+B2=H242+32=H2H2=16+9H2=25H=5Therefore,sinA=PH=45cosA=BH=35Now,1-sinA1+cosA=1-451+35                  =5-455+35                  =1585                  =18                  =122Hence, 1-sinA1+cosA=122.

Page No 547:

Question 21:

If cot θ=158 then evaluate 1+sin θ 1-sin θ1+cos θ 1-cos θ.

Answer:

Given: cotθ=158Since, cotθ=BPP=8 and B=15Using Pythagoras theorem,P2+B2=H282+152=H2H2=64+225H2=289H=17Therefore,sinθ=PH=817cosθ=BH=1517Now,1+sinθ1-sinθ1+cosθ1-cosθ=1-sin2θ1-cos2θ                               =cos2θsin2θ             sin2θ+cos2θ=1                               =cot2θ                               =1582                               =22564Hence, 1+sinθ1-sinθ1+cosθ1-cosθ=22564.

Page No 547:

Question 22:

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 547:

Question 23:

In the given figure, ABCD is a rectangle in which diag. AC = 17 cm, ∠BCA = θ and sin θ=817.
Find (i) the area of rect. ABCD, (ii) the perimeter of rect. ABCD.

Answer:

Given: In ABC,AC=17 cmsinθ=817Since, sinθ=PHP=8 and H=17Using Pythagoras theorem,P2+B2=H282+B2=172B2=289-64B2=225B=15Therefore,AB= 8 cm and BC=15 cmTherefore,i Area of rectangle ABCD=AB×BC                                       =8×15                                       =120 cm2ii Perimeter of rectangle ABCD=2AB+BC                                               =28+15                                               =223                                               =46 cm

Page No 547:

Question 24:

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 548:

Question 25:

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

Page No 548:

Question 26:

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 548:

Question 27:

If cot A+1cot A=2, find the value of cot2A+1cot2A.

Answer:

Given:cotA+1cotA=2cotA+1cotA=2Squaring both sides, we getcotA+1cotA2=22cot2A+1cotA2+2cotA1cotA=4cot2A+1cot2A+2=4cot2A+1cot2A=4-2cot2A+1cot2A=2Hence, the value of cot2A+1cot2A is 2.

Page No 548:

Question 28:

If 3 tan θ=3 sin θ, find the value of sin θ.

Answer:

Given:3tanθ=3sinθ3tanθ=3sinθ3sinθcosθ=3sinθ3sinθcosθ-3sinθ=03sinθ-3sinθcosθcosθ=03sinθ-3sinθcosθ=03sinθ1-3cosθ=0sinθ=0 or 1-3cosθ=0sinθ=0 or cosθ=13sinθ=0 or cos2θ=13sinθ=0 or 1-cos2θ=1-13sinθ=0 or sin2θ=23sinθ=0 or sinθ=23Hence, the value of sinθ is 0 or 23.

Page No 548:

Question 29:

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 548:

Question 30:

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 555:

Question 1:

If tan θ=815 then cosec θ=?
(a) 1517

(b) 1715

(c) 178

(d) 817

Answer:

Given: tanθ=815Since, tanθ=PBP=8 and B=15Using Pythagoras theorem,P2+B2=H282+152=H2H2=64+225H2=289H=17Therefore,cosecθ=HP=178

Hence, the correct option is (c).

Page No 555:

Question 2:

If tan θ=3 then sec θ = ?

(a) 2

(b) 12

(c) 32

(d) 23

Answer:

Given: tanθ=31Since, tanθ=PBP=3 and B=1Using Pythagoras theorem,P2+B2=H232+12=H2H2=3+1H2=4H=2Therefore,secθ=HB=21=2

Hence, the correct option is (a).

Page No 555:

Question 3:

If cosec θ=10 then sec θ=?

(a) 110

(b) 210

(c) 310

(d) 103

Answer:

Given: cosecθ=101Since, cosecθ=HPP=1 and H=10Using Pythagoras theorem,P2+B2=H212+B2=102B2=10-1B2=9B=3Therefore,secθ=HB=103

Hence, the correct option is (d).

Page No 555:

Question 4:

If sec θ=257 then sin θ=?

(a) 725

(b) 2425

(c) 724

(d) 247

Answer:

Given: secθ=257Since, secθ=HBB=7 and H=25Using Pythagoras theorem,P2+B2=H2P2+72=252P2=625-49P2=576P=24Therefore,sinθ=PH=2425

Hence, the correct option is (b).

Page No 555:

Question 5:

If sin θ=12 then cot θ=?
(a) 32

(b) 1

(c) 3

(d) 13

Answer:

Given: sinθ=12Since, sinθ=PHP=1 and H=2Using Pythagoras theorem,P2+B2=H212+B2=22B2=4-1B2=3B=3Therefore,cotθ=BP=31

Hence, the correct option is (c).

Page No 555:

Question 6:

If cos θ=45 then tan θ=?
(a) 34

(b) 43

(c) 35

(d) 53

Answer:

Given: cosθ=45Since, cosθ=BHB=4 and H=5Using Pythagoras theorem,P2+B2=H2P2+42=52P2=25-16P2=9P=3Therefore,tanθ=PB=34

Hence, the correct option is (a).



Page No 556:

Question 7:

If tan θ=43 then sin θ+cos θ=?
(a) 73

(b) 74

(c) 75

(d) 57

Answer:

Given: tanθ=43Since, tanθ=PBP=4 and B=3Using Pythagoras theorem,P2+B2=H242+32=H2H2=16+9H2=25H=5Therefore,sinθ=PH=45cosθ=BH=35Now,sinθ+cosθ=45+35                 =75

Hence, the correct option is (c).

Page No 556:

Question 8:

If (tan θ + cot θ) = 5 then (tan2 θ + cot2 θ) = ?
(a) 27
(b) 25
(c) 24
(d) 23

Answer:

Given:tanθ+cotθ=5tanθ+cotθ=5Squaring both sides, we gettanθ+cotθ2=52tan2θ+cot2θ+2cotθtanθ=25tan2θ+cot2θ+21tanθtanθ=25           cotθ=1tanθtan2θ+cot2θ+2=25tan2θ+cot2θ=23Hence, the correct option is d.

Page No 556:

Question 9:

If (cos θ + sec θ) = 52 then (cos2 θ + sec2 θ) = ?
(a) 174

(b) 214

(c) 294

(d) 334

Answer:

Given:cosθ+secθ=52cosθ+secθ=52Squaring both sides, we getcosθ+secθ2=522cos2θ+sec2θ+2cosθsecθ=254cos2θ+sec2θ+2cosθ1cosθ=254            secθ=1cosθcos2θ+sec2θ+2=254cos2θ+sec2θ=254-2cos2θ+sec2θ=25-84cos2θ+sec2θ=174Hence, the correct option is a.

Page No 556:

Question 10:

If 4tan θ = 3 then (cos2 θ – sin2 θ) = ?
(a) 425

(b) 725

(c) 1

(d) 1125

Answer:

Given: 4tanθ=3tanθ=34Since, tanθ=PBP=3 and B=4Using Pythagoras theorem,P2+B2=H232+42=H2H2=9+16H2=25H=5Therefore,sinθ=PH=35cosθ=BH=45cos2θ-sin2θ=452-352                   =1625-925                   =16-925                   =725Hence, the correct option is b.

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Question 11:

If 4cot θ=3 then sin θ-cos θsin θ+cos θ=?
(a) 37

(b) 27

(c) 17

(d) 0

Answer:

Given: 4cotθ=3cotθ=34Since, cotθ=BPP=4 and B=3Using Pythagoras theorem,P2+B2=H242+32=H2H2=16+9H2=25H=5Therefore,sinθ=PH=45cosθ=BH=35sinθ-cosθsinθ+cosθ=45-3545+35                     =4-354+35                     =17Hence, the correct option is c.

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Question 12:

If 3cos θ = 2 then (2sec2 θ + 2tan2 θ – 7) = ?
(a) 0
(b) 1
(c) 3
(4) 4

Answer:

Given: 3cosθ=2cosθ=23Since, cosθ=BHB=2 and H=3Using Pythagoras theorem,P2+B2=H2P2+22=32P2=9-4P2=5P=5Therefore,secθ=HB=32tanθ=PB=52Now,2sec2θ+2tan2θ-7=2322+2522-7                               =294+254-7                               =92+52-7                               =9+52-7                               =142-7                               =7-7                               =0Hence, the correct option is a.

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Question 13:

If sec θ + tan θ + 1 = 0 then (sec θ – tan θ) = ?
(a) 1
(b) –1
(c) 0
(d) 2

Answer:

Given: secθ+tanθ+1=0secθ+tanθ+1=0secθ+tanθ=-1Multiplying and dividing LHS by secθ-tanθ, we getsecθ+tanθ×secθ-tanθsecθ-tanθ=-1sec2θ-tan2θsecθ-tanθ=-11+tan2θ-tan2θsecθ-tanθ=-1           sec2θ=1+tan2θ1secθ-tanθ=-1secθ-tanθ=-1Hence, the correct option is b.

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Question 14:

If cos A + cos2 A = 1 then (sin2 A + sin4 A) = ?
(a) 12

(b) 2

(c) 1

(d) 4

Answer:

Given: cosA+cos2A=1cosA+cos2A=1cosA=1-cos2AcosA=sin2A          sin2A+cos2A=1Now,sin2A+sin4A=sin2A+sin2A2                   =cosA+cosA2                   =cosA+cos2A                   =1Hence, the correct option is c.

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Question 15:

If sin θ=32then cosec θ+cot θ=?

(a) 2+3

(b) 23

(c) 2

(d) 3

Answer:

Given: sinθ=32Since, sinθ=PHP=3 and H=2Using Pythagoras theorem,P2+B2=H232+B2=22B2=4-3B2=1B=1Therefore,cosecθ=HP=23cotθ=BP=13Now,cosecθ+cotθ=23+13                      =2+13                      =33                      =3Hence, the correct option is d.

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Question 16:

If 3 tan θ = 3sin θ then (sin2θ – cos2θ) = ?

(a) 13

(b) 13

(c) 3

(d) 23

Answer:

Given:3tanθ=3sinθ3tanθ=3sinθ3sinθcosθ=3sinθ3sinθcosθ-3sinθ=03sinθ-3sinθcosθcosθ=03sinθ-3sinθcosθ=03sinθ1-3cosθ=0sinθ=0 or 1-3cosθ=0sinθ=0 or cosθ=13sinθ=0 or cos2θ=13sinθ=0 or 1-cos2θ=1-13sinθ=0 or sin2θ=23sinθ=0 or sinθ=23For sinθ=0,sin2θ=01-sin2θ=1-0cos2θ=1Thus, sin2θ-cos2θ=-1For sinθ=23,sin2θ=231-sin2θ=1-23cos2θ=13Thus, sin2θ-cos2θ=23-13=13 

Hence, the correct option is (a).
 



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