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Trigonometric Ratios And Trigonometric Identities

Class 10th Mathematics West Bengal Board Solution
Let Us Work Out 23.1
  1. I have drawn a right angled triangle ABC whose hypotenuse AB = 10 cm., base BC = 8 cm.…
  2. Soma has drawn a right angled triangle ABC whose ABC = 90, AB = 24 cm. and BC = 7 cm.…
  3. If in a right angled triangle ABC, ∠C=90°, DC = 21 unit and AB = 29 units, then let us…
  4. If costheta = 7/25 cos then let us determine the values of all trigonometric ratio of…
  5. If cot θ = 2, then let us determine the values of tan θ and sec θ and show that 1+…
  6. If cos θ = 0.6, then let us show that, (5 sin θ - 3 tan θ) = 0.
  7. If cota = 4/7.5 cot then let us determine the values of cos A and cosec A and show that…
  8. If sinc = 2/3 then let us write by calculating, the value of cos C × cosec C.…
  9. Let us write with reason whether the following statements are true or false : i. The…
Let Us Work Out 23.2
  1. In the window of our house, there is a ladder at an angle of 60o with the ground. If…
  2. ABC is a right angled triangle with its ∠B is 1 right angle. If AB = 8√3 cm. and BC = 8…
  3. In a right angled triangle ABC, ∠B=90°, ∠A=30° and AC = 20 cm. Let us determine the…
  4. In a right angled triangle PQR, ∠Q=90°, ∠R=45°; if PR = 3√2, then let us find out the…
  5. sin^2 45° - cosec^2 60° + sec^2 30° Let us determine the values of :…
  6. sec^2 45° - cot^2 45° + sin^2 30° - sin^2 60° Let us determine the values of :…
  7. 3tan^245^circle - sin^260^circle - 1/3 cot^230^circle - 1/8 sec^245^circle Let us…
  8. 4/3 cot^230^circle + 3sin^260^circle - 2cosec^260^circle - 3/4 tan^230^circle Let us…
  9. 1/3 cos30^circle / 1/2 sin45^circle + tan60^circle /cos30^circle Let us determine the…
  10. cot^230^circle - 2cos^260^circle - 3/4 sec^245^circle - 4 sin 30° Let us determine the…
  11. sec^260^circle - cot^230^circle - 2tan30^circle cosec60^circle /1+tan^230^circle Let…
  12. tan60^circle - tan30^circle /1+tan60^circle tan30^circle + cos60^circle cos30^circle +…
  13. 1-sin^230^circle /1+sin^230^circle x cos^260^circle + cos^230^circle /cosec^290^circle…
  14. sin^2 45° + cos^2 45° = 1 Let us show that,
  15. cos60° = cos^2 30° - sin^2 30° Let us show that,
  16. 2tan30^circle /1-tan^230^circle = root 3 Let us show that,
  17. root 1+cos30^circle /1-cos30^circle = sec60^circle + tan60^circle Let us show that,…
  18. 2tan^230^circle /1-tan^230^circle + sec^245^circle - cot^245^circle = sec 60° Let us…
  19. tan^2 pi /4 sin pi /3 tan pi /6 tan^2 pi /3 = 1 1/2 Let us show that,…
  20. sin pi /3 tan pi /6 + sin pi /2 cos pi /3 = 2sin^2 pi /4 Let us show that,…
  21. If x sin 45° cos 45° tan 60°= tan^2 45° - cos 60°, then let us determine the value of…
  22. If xsin60^circle cos^230^circle = tan^245^circle sec60^circle /cosec60^circle then let…
  23. If x^2 = sin^2 30° + 4 cot^2 45° - sec^2 60°, then let us determine the value of x.…
  24. If x tan 30° + y cot 60° = 0 and 2x - y tan 45° = 1, then let us write, by calculating…
  25. sin (A+B) = sin A cos B + cos A sin B If A = B = 45°, then let us justify…
  26. cos (A+B) = cos A cos B - sin A sin B If A = B = 45°, then let us justify…
  27. In an equilateral triangle ABC, BD is a median. Let us prove that, ∠ ABD = cot ∠BAD…
  28. In an isosceles triangle ABC, AB = AC and ∠BAC = 90°; the bisector of ∠BAC intersects…
  29. Let us determine the value/value of for which 2 cos^2 θ - 3 cos θ + 1 = 0 will be…
Let Us Work Out 23.3
  1. If sintegrate heta = 4/5 then let us write the value of cosectheta /1+cottheta by…
  2. If tantheta = 3/4 then let us show that root 1-sintegrate heta /1+sintegrate heta =…
  3. If tan θ=1, then let us determine the value of 8sintegrate heta +5costheta /sin^3theta…
  4. i. Let us express cosec θ and tan θ in term of sin θ. ii. Let us write cosec θ and tan…
  5. If sec θ + tan θ=2, then let us determine the value of (sec θ-tan θ).…
  6. If cosec θ - cot θ = √2 - 1, then let us write by calculating, the value of (cosec θ +…
  7. If sin θ + cos θ = 1, then let us determine the value of sin θ × cos θ.…
  8. If tan θ × cot θ = 2, then let us determine the value of (tan θ - cot θ).…
  9. If sintegrate heta -costheta = 7/13 then let us determine the value of sin θ + cos θ.…
  10. If sintegrate heta costheta = 1/2 then let us write by calculating, the value of (sin…
  11. If sectheta -tantheta = 1/root 3 then let us determine the values of both sec θ and…
  12. If cosec θ + cot θ = √3, then let us determine the values of both cosec θ and cot θ.…
  13. If sintegrate heta +costheta /sintegrate heta -costheta = 7 then let us write by…
  14. If cosectheta +sintegrate heta /cosectheta -sintegrate heta = 5/3 then let us write by…
  15. If sectheta +costheta = 5/3 then let us write by calculating, the value of (sec θ-cos…
  16. Let us determine the value of from the relation 5sin^2theta +4cos^2theta = 9/2…
  17. If tan^2theta +cot^2theta = 10/3 then let us determine the values of tan θ + cot θ and…
  18. If sec^2theta +tan^2theta = 13/12 then let us write by calculating, the value of…
  19. In ΔPQR, ∠Q is right angle. If PR = √5 units and PQ - RQ = 1 unit, then let us…
  20. In ΔXYZ, ∠Y is right angle. If XY = 2√3 units and XZ - YZ = 2 units then let us…
  21. x = 2 sin θ, y = 3 cos θ Let us eliminate ‘θ’ from the relations :…
  22. 5x = 3 sec θ, y = 3 tan θ Let us eliminate ‘θ’ from the relations :…
  23. If sinalpha = 5/13 then let us show that, tan α + sec α = 1.5.
  24. If tana = n/m then let us determine the values of both sinA and secA.…
  25. If costheta = x/root x^2 + y^2 then let us show that, x sin θ = y cos θ.…
  26. If sinalpha = a^2 - b^2/a^2 + b^2 then let us show that, cotalpha = 2ab/a^2 - b^2…
  27. If sintegrate heta /x = costheta /y then let us show that, sintegrate heta -costheta =…
  28. If (1+4x^2) cos A = 4x, then let us show that, coseca+cota = 1+2x/1-2x…
  29. If x = a sin θ and y = b tan θ, then let us prove that, a^2/x^2 - b^2/y^2 = 1…
  30. If sin θ + sin^2 θ = 1, then let us prove that, cos^2 θ + cos^4 θ = 1.…
  31. Q9A1 If 3x = cosec α and 3/x = cotalpha then the value of 3 (x^2 - 1/x^2) isA. 1/27 B.…
  32. Q9A2 If 2x = sec A and 2/x = tana then the value of 2 (x^2 - 1/x^2) isA. 1/2 B. 1/4 C. 1/8…
  33. Q9A3 If tan α + cot α = 2, then the value of (tan^13 α + cot^13 α) isA. 1 B. 0 C. 2 D.…
  34. Q9A4 If sintegrate heta -costheta = 0 (0^circle less than equal to theta less than equal…
  35. Q9A5 If 2 cos 3 θ = 1, then the value of θ isA. 10o B. 15o C. 20o D. 30o…
  36. Let us write whether the following statements are true or false : i. If 0^circle less…
  37. Let us fill in the blanks : i. The value of (4/sec^2theta + 1/1+cot^2theta…
  38. If and then let us determine the values of both r and θ.
  39. If sin A + sin B = 2 where 0^circle less than equal to a less than equal to 90^circle…
  40. If 0° θ 90°, then let us calculate the least value of (9 tan^2 θ + 4 cot^2 θ).…
  41. Let us calculate the value of (sin^2 α + cos^6 α + 3 sin^2 α cos^2 α).…
  42. If cosec^2 θ = 2 cot θ and 0°θ90°, then let us determine the value of θ.…

Let Us Work Out 23.1
Question 1.

I have drawn a right angled triangle ABC whose hypotenuse AB = 10 cm., base BC = 8 cm. and perpendicular AC = 6 cm. Let us determine the values of sine and tangent ∠ABC.


Answer:

Given, a right angled triangle with AB = 10cm(hypothesis), BC = 8cm (base) and AC = 6cm (perpendicular)


Need to find out the value of sinθ and tanθ


⇒ now, we know that


And



=


= 0.6 cm



=


= 0.75cm



Question 2.

Soma has drawn a right angled triangle ABC whose ABC = 90, AB = 24 cm. and BC = 7 cm. By calculating, let us write the value of sinA, cosA, tanA and cosec A.



Answer:

Given, AB = 24, BC = 7


Need to find the values of sinA, cosA, tanA and cosecA


⇒ [from the figure, AC is hypothesis, AB is base and CB is perpendicular]


⇒ AC2 = AB2 + BC2


⇒ AC2 = (24)2 + (7)2


= 576 + 49


= 625


⇒ AC2 = 625


⇒ AC = 25



=


=


= 0.28


⇒ sinA = 0.28



=


=


= 0.96


⇒ cosA = 0.96



=


=


= 0.29


⇒ tanA = 0.29



=


=


= 3.57


⇒ cosecA = 3.57



Question 3.

If in a right angled triangle ABC, ∠C=90°, DC = 21 unit and AB = 29 units, then let us find the values of sinA, cosA, sinB and cosB.


Answer:

Given, BC = 21 and AB = 29


Need to find the values of sinA, cosA,sinB,cosB


and


⇒ AB2 = AC2 + BC2


⇒ (29)2 = AC2 + (21)2


⇒ 841 = AC2 + 441


⇒ 400 = AC2


⇒ AC = 20



= = 0.72



= = 0.68



= = 0.68



= = 0.72



Question 4.

If cos then let us determine the values of all trigonometric ratio of the angle θ.


Answer:

Given,


Need to find the trigonometric ratios



[we know that hypothesis2 = perpindicular2 + base2]


⇒ (25)2 = P2 + (7)2


⇒ 625 = P2 + 49


⇒ 576 = P2


⇒ P = 24



= = 0.96



= = 0.28



= = 3.42



= = 0.291



= = 1.04



= = 3.57



Question 5.

If cot θ = 2, then let us determine the values of tan θ and sec θ and show that 1+ tan2 θ = sec2 θ.


Answer:

Given, cotθ = 2


Need to show 1 + tan2θ = sec2θ


= 2


=


⇒ hypothesis2 = perpindicular2 + base2


= 1 + 22


= 5


⇒ hypothesis = √5


=


⇒ 1 + tan2θ =


=


=


⇒ sec2θ =


=


∴ LHS = RHS



Question 6.

If cos θ = 0.6, then let us show that, (5 sin θ – 3 tan θ) = 0.


Answer:

Given, cosθ = 0.6



= 0.6 = =


[we know that hypothesis2 = perpindicular2 + base2]


⇒ (5)2 = P2 + 32


⇒ 25 = P2 + 9


⇒ P = 4



=



=


⇒ 5sinθ – 3tanθ


=


= 4-4


= 0


Hence, proved



Question 7.

If cot then let us determine the values of cos A and cosec A and show that 1 + cot2 A = cosec2A.


Answer:

Given,



=


[we know that hypothesis2 = perpindicular2 + base2]


⇒ h2 = 7.52 + 42


= 72.25


⇒ h = 8.5



=



=


⇒ 1 + cot2A =


=


=


= 1.28


⇒ cosec2A =


=


= 1.28


∴ 1 + cot2A = cosec2A



Question 8.

If then let us write by calculating, the value of cos C × cosec C.


Answer:

Given, sinC =



=


[we know that hypothesis2 = perpindicular2 + base2]


⇒ 32 = 22 + b2


⇒ 9 = 4 + b2


⇒ b = √5



=



=


⇒ cosC × cosecC = ×


=



Question 9.

Let us write with reason whether the following statements are true or false :

i. The value of tan A is always greater than 1.

ii. The value of cot A is always less than 1.

iii. For an angle θ, it may be possible that

iv. For an angle α, it may be possible that,

v. For an angle β (Beta) it may be possible that,

vi. For an angle θ, it may be possible that,


Answer:

(i) TRUE


Let us consider ABC as a right angled triangle where ∠B = 90°



=


Let us assume that BC is greater than AB


Ex: BC = 6 and AB = 2



= 3


∴ tanA > 1


(ii) FALSE


Let us consider ABC as a right angled triangle where ∠B = 90°



=


Let us assume that AB is greater than BC


Ex: BC = 2 and AB = 8



= 4


∴ cotA > 1


(iii) FALSE


Given,


⇒ sinθ = 1.33


⇒ From the trigonometric ratios we know that sinθ value must lie between 0° to 90°


And sin90 = 1


(iv) FALSE


Given,


⇒ secα = 2.4


⇒ 2.4 does not lie between 0 and 2


(v) TRUE


Given,


⇒ cosecβ = 0.38


⇒ The value of cosec lie between 2 and 1


(vi) TRUE


Given,


⇒ cosθ = 0.6


⇒ The value of cos lie between 0 and 1, i.e 0° to 90°




Let Us Work Out 23.2
Question 1.

In the window of our house, there is a ladder at an angle of 60o with the ground. If the ladder is 2√3m. long, then let us write by calculating with figure, the height of our window above the ground.


Answer:

Given, length of ladder is 2√3 m


Angle is 60°



⇒ ∠C = 60° and AC = 2√3


Need to find out AB


⇒ we know, sin60 =




⇒ P =


= 3


∴ the height of the window = 3m



Question 2.

ABC is a right angled triangle with its ∠B is 1 right angle. If AB = 8√3 cm. and
BC = 8 cm., then let us write by calculating, the values of ∠ACB and ∠BAC.


Answer:

Given, ∠ B = 90°


Need to find ∠ A and ∠ C


⇒ from the figure ∠ ACB,




Question 3.

In a right angled triangle ABC, ∠B=90°, ∠A=30° and AC = 20 cm. Let us determine the lengths of two sides BC and AB.


Answer:

Given, ∠ B = 90° and ∠ A = 30°


AC = 20



Question 4.

In a right angled triangle PQR, ∠Q=90°, ∠R=45°; if PR = 3√2, then let us find out the lengths of two sides PQ and QR.


Answer:


To Find: PQ and QR


Given: PR = 3√2 and ∠Q=90°, ∠R=45°





PQ = 3


Now, we also know that


cos θ =




QR = 3


Hence PQ = 3 units and QR = 3 units



Question 5.

Let us determine the values of :

sin245° – cosec260° + sec230°


Answer:




⇒ sin245 – cosec260 + sec230


=


=


=



Question 6.

Let us determine the values of :

sec245° – cot245° + sin230° – sin260°


Answer:

⇒ sec45 = √2


⇒ cot45 = 1




⇒ sec245 – cot245 –sin230 –sin260


=


=


=


= 0



Question 7.

Let us determine the values of :



Answer:

⇒ tan45 = 1






=


=


=


= 1



Question 8.

Let us determine the values of :



Answer:






=


=


=


=



Question 9.

Let us determine the values of :



Answer:





=


=


=



Question 10.

Let us determine the values of :

sin 30°


Answer:



⇒ sec45 = √2




=


=


=


=



Question 11.

Let us determine the values of :



Answer:




⇒ sec60 = 2



=


= = 1


⇒ sec260 – cot230 -1


= 22-√32 -1


= 0



Question 12.

Let us determine the values of :

sin 60°sin 30°


Answer:








=


=


=



=


=


=



Question 13.

Let us determine the values of :

(sin60° tan30°)


Answer:




⇒ cosec90 = 1


⇒ cot90 = 0





=


=


=


=



Question 14.

Let us show that,

sin245° + cos245° = 1


Answer:



⇒ sin245 + cos245


=


=


= 1



Question 15.

Let us show that,

cos60° = cos230° – sin230°


Answer:




⇒ cos230-sin230


=


=


=


=


= cos60



Question 16.

Let us show that,



Answer:



=


=


=



Question 17.

Let us show that,



Answer:


⇒ sec60 = 2




=


=



Question 18.

Let us show that,

= sec 60°


Answer:


⇒ sec45 = √2


⇒ cot45 = 1


⇒ sec60 = 2



=


=



=


=


=



Question 19.

Let us show that,



Answer:






=


=



Question 20.

Let us show that,



Answer:







=


=


= 1



=


= 1



Question 21.

If x sin 45° cos 45° tan 60°= tan2 45° - cos 60°, then let us determine the value of x.


Answer:





⇒ tan45 = 1


⇒ xsin45 cos45 tan60 = tan245-cos60


=





Question 22.

If then let us determine the value of x.


Answer:



⇒ tan45 = 1




⇒ xsin60 cos230 =






Question 23.

If x2 = sin2 30° + 4 cot2 45° - sec2 60°, then let us determine the value of x.


Answer:


⇒ cot45 = 1



⇒ x2 = sin230 + 4cot245-sec260







Question 24.

If x tan 30° + y cot 60° = 0 and 2x – y tan 45° = 1, then let us write, by calculating the values of x and y.


Answer:

⇒ tan45 = 1




⇒ xtan30 + ycot60 = 0



= 0


⇒ x + y = 0


⇒ x = -y


⇒ 2x-ytan45 = 1


⇒ 2(-y)-y(1) = 1


⇒ -3y = 1





Question 25.

If A = B = 45°, then let us justify

sin (A+B) = sin A cos B + cos A sin B


Answer:

Given, A = B = 45


⇒ A + B = 90




⇒ sin90 = 1


⇒ sin90 = sin45 cos45 + cos45 sin45




⇒ 1 = 1



Question 26.

If A = B = 45°, then let us justify

cos (A+B) = cos A cos B – sin A sin B


Answer:

Given, A = B = 45


⇒ A + B = 90




⇒ cos90 = 0


⇒ cos(A + B) = cosA cosB-sinA sinB


⇒ cos90 = cos45 cos45-sin45 sin45



⇒ 0 = 0



Question 27.

In an equilateral triangle ABC, BD is a median. Let us prove that, ∠ ABD = cot ∠BAD


Answer:


To Prove : cot ∠ ABD = cot ∠BAD


Proof:


In the figure shown above ABC is an equilateral triangle


Now BD is a median on side AC, Therefore, AD = DC


And ∠ ADB = 90°


(By property of equilateral triangle if a median is dropped from one vertex to opposite side, it is perpendicular to the side, and ∠ ABD = ∠BAD = 45°)


If the angles are equal, their values of cot will also be equal


So,


Cot ∠ ABD = cot ∠BAD


Hence, Proved



Question 28.

In an isosceles triangle ABC, AB = AC and ∠BAC = 90°; the bisector of ∠BAC intersects the side BC at the point D.

Let us prove that,


Answer:


To Prove:


Proof:


Since, ABC is an isosceles triangle with AB = AC then by the property of isosceles triangle that the angles opposite to equal sides are also equal, we have


∠ABC = ∠ACB


From Triangle ABC,


Sum of angles of triangle = 180°


Therefore,


∠ABC + ∠BAC + ∠ACB = 180°


2 ∠ABC + 90° = 180°


∠ABC = 45°


And ∠ACB = 45°


Sec ∠ACD = sec 45° =


∠CAD = 45° (∠CAD = ∠ACB)


Sin ∠CAD =




Hence,


= 2


Hence, Proved.



Question 29.

Let us determine the value/value of for which 2 cos2 θ – 3 cos θ + 1 = 0 will be true.


Answer:

Given, 2cos2θ -3cosθ + 1


Need to check the equation is true for different values of theta


⇒ let θ = 0°


⇒ 2cos20° -3cos0° + 1


⇒ cos0° = 1


⇒ 2(1)2 – 3(1) + 1


= 2-3 + 1


= -1 + 1


= 0


Hence, if θ is 0 the given equation will be true




Let Us Work Out 23.3
Question 1.

If then let us write the value of by determining it.


Answer:

Given,


⇒ sin2θ + cos2θ = 1











=


=



Question 2.

If then let us show that


Answer:

Given,


⇒ sec2θ = 1 + tan2θ



=


=






⇒ cos θ =


⇒ sin2θ + cos2θ = 1



=


=




=


=


=



Question 3.

If tan θ=1, then let us determine the value of


Answer:

Given, tan θ = 1




⇒ sinθ = cosθ


⇒ sec2θ = 1 + tan2θ



⇒ secθ = √2





=


=


=


=


=


=



Question 4.

i. Let us express cosec θ and tan θ in term of sin θ.

ii. Let us write cosec θ and tan θ in term cos θ.


Answer:

(i) ⇒




(ii) ⇒



=




Question 5.

If sec θ + tan θ=2, then let us determine the value of (sec θ-tan θ).


Answer:

Given, secθ + tanθ = 2


⇒ sec2θ = 1 + tan2θ


⇒ sec2θ - tan2θ = 1


⇒ (secθ + tanθ)( secθ- tanθ) = 1


⇒ 2(secθ- tanθ) = 1




Question 6.

If cosec θ – cot θ = √2 – 1, then let us write by calculating, the value of (cosec θ + cot θ).


Answer:

Given, cosecθ –cotθ = √2 -1


⇒ cosec2θ = 1 + cot2θ


⇒ cosec2θ - cot2θ = 1


⇒(cosecθ + cotθ)(cosecθ – cotθ) = 1


⇒(cosecθ + cotθ)(√2-1) = 1




Question 7.

If sin θ + cos θ = 1, then let us determine the value of sin θ × cos θ.


Answer:

Given, sinθ + cosθ = 1


⇒ squaring on both sides


⇒ (sinθ + cosθ)2 = 12


⇒ sin2θ + cos2θ + 2sinθcosθ = 1


⇒ 1 + 2sinθ cosθ = 1


⇒ 2sinθ cosθ = 0


⇒ sinθ × cosθ = 0



Question 8.

If tan θ × cot θ = 2, then let us determine the value of (tan θ - cot θ).


Answer:

Given, tanθ + cotθ = 2




⇒ tan2θ -2tanθ + 1 = 0


⇒ (tanθ -1)(tanθ -1) = 0


∴ tanθ = 1


= 1


⇒ tanθ –cotθ = 0



Question 9.

If then let us determine the value of sin θ + cos θ.


Answer:

Given,


⇒ squaring on both sides







⇒ sin2θ + cos2 θ = 1, Can be written as


⇒ (sinθ + cosθ)2 – 2sinθcosθ = 1








Question 10.

If then let us write by calculating, the value of (sin θ + cos θ).


Answer:


⇒ 2sinθcosθ = 1


⇒ we know that, sin2θ + cos2θ = 1


Can be written as


⇒ (sinθ + cosθ)2 – 2sinθcosθ = 1


⇒ (sinθ + cosθ)2-1 = 1


⇒ (sinθ + cosθ)2 = 2


⇒ (sinθ + cosθ) = √2



Question 11.

If then let us determine the values of both sec θ and tan θ.


Answer:

Given,


-----eq(1)


⇒ sec2θ –tan2θ = 1


⇒ (secθ + tanθ)(secθ –tanθ) = 1


⇒ (


⇒ secθ + tanθ = √3………eq(2)


From adding both the equations we get










Question 12.

If cosec θ + cot θ = √3, then let us determine the values of both cosec θ and cot θ.


Answer:

Given, cosecθ + cotθ = √3


⇒ cosecθ + cotθ = √3 ……..eq(1)


⇒ cosec2θ –cot2θ = 1


⇒ (cosecθ + cotθ)(cosecθ –cotθ) = 1


⇒ (√3)(cosecθ –cotθ) = 1


………eq(2)


From adding both the equations we get










Question 13.

If then let us write by calculating, the value of tan θ.


Answer:

Given,


can be written as


[by compendo and dividend rule]





Question 14.

If then let us write by calculating, the value of sin θ.


Answer:

Given,











Question 15.

If then let us write by calculating, the value of (sec θ-cos θ).


Answer:

Given,




⇒ 3sec2θ -5secθ + 3 = 0



Question 16.

Let us determine the value of from the relation


Answer:

Given,


Need to find tantθ value


⇒ sin2θ + cos2θ = 1












=


⇒ tanθ = 1



Question 17.

If then let us determine the values of tan θ + cot θ and tan θ – cot θ and from these let us write the value of tan θ.


Answer:

Given,




⇒ 3tan4θ + 3-10tan2θ = 0


⇒ (3tan2θ -1)(tan2θ -3) = 0


⇒ tanθ = √3 and


and


⇒ tanθ + cotθ =




Question 18.

If then let us write by calculating, the value of (sec4 – tan4 θ).


Answer:

Given,


⇒ sec2θ = 1 + tan2θ








⇒ (sec4θ –tan4θ)


= (sec2θ)2-(tan2θ)2


=


=


=



Question 19.

In ΔPQR, ∠Q is right angle. If PR = √5 units and PQ – RQ = 1 unit, then let us determine the value of cosP – cosR.


Answer:

Given, ∠ Q = 90°


PR = √5 and PQ –QR = 1


⇒ cosP –cosR


=


=


=



Question 20.

In ΔXYZ, ∠Y is right angle. If XY = 2√3 units and XZ – YZ = 2 units then let us determine the values of (secX - tanX).


Answer:

Given, ∠ Y = 90°


XY = 2√3 and XZ-YZ = 2


⇒ secX-tanX


=


=


=



Question 21.

Let us eliminate ‘θ’ from the relations :

x = 2 sin θ, y = 3 cos θ


Answer:

Given, x = 2sinθ and y = 3cosθ




⇒ sin2θ + cos2θ = 1





Question 22.

Let us eliminate ‘θ’ from the relations :

5x = 3 sec θ, y = 3 tan θ


Answer:

Given, 5x = 3secθ and y = 3tanθ


and


⇒ sec2θ –tan2θ = 1





Question 23.

If then let us show that, tan α + sec α = 1.5.


Answer:

Given,


⇒ sin2α + cos2α = 1














Question 24.

If then let us determine the values of both sinA and secA.


Answer:

Given,



⇒ sec2A = 1 + tan2A


=


=




=


=




Question 25.

If then let us show that, x sin θ = y cos θ.


Answer:

Given,


⇒ sin2θ + cos2θ = 1


⇒ sin2θ = 1-cos2θ


=


=









⇒ xsinθ = ycosθ



Question 26.

If then let us show that,


Answer:

Given,


⇒ sin2α + cos2α = 1


⇒ cos2α = 1-sin2α


=


=


=




=




Question 27.

If then let us show that,


Answer:

Given,



⇒ sin2θ + cos2θ = 1










=


=




Question 28.

If (1+4x2) cos A = 4x, then let us show that,


Answer:

Given,


⇒ sin2A + cos2A = 1


⇒ sin2A = 1-cos2A


=


=


=


=




⇒ cosecA + cotA


=


=


=


=


=


=



Question 29.

If x = a sin θ and y = b tan θ, then let us prove that,


Answer:

Given, x = asinθ and y = btanθ










⇒ sec2θ – tan2θ = 1







Question 30.

If sin θ + sin2 θ = 1, then let us prove that, cos2 θ + cos4 θ = 1.


Answer:

Given, sinθ + sin2θ = 1


⇒ sinθ + sin2θ = sin2θ + cos2θ


⇒ sinθ = cos2θ


⇒ squaring on both sides


⇒ (sinθ)2 = (cos2θ)2


⇒ sin2θ = cos4θ


⇒ sinθ + sin2θ = 1


⇒ cos2θ + cos4θ = 1



Question 31.

If 3x = cosec α and then the value of is
A.

B.

C.

D.


Answer:

Given, 3x = cosecα and




=


=


=


⇒ option A is incorrect as it does not satisfy the given equation


⇒ option B is incorrect as it does not satisfy the given equation


⇒ option D is incorrect as it does not satisfy the given equation


Question 32.

If 2x = sec A and then the value of is
A.

B.

C.

D.


Answer:

⇒ Given, 2x = secA and




=


=


=


⇒ Option B is incorrect as it does not satisfy the value


⇒ Option C is incorrect as it does not satisfy the value


⇒ Option D is incorrect as it does not satisfy the value


Question 33.

If tan α + cot α = 2, then the value of (tan13 α + cot13 α) is
A. 1

B. 0

C. 2

D. none of these


Answer:

⇒ tanα + cotα = 2




⇒ 1 + tan2α -2tanα = 0


⇒ (tanα -1)2 = 0


⇒ tanα = 1


⇒ cotα = 1


⇒ tan13α + cot13α = 1 + 1 = 2


⇒ Option A is incorrect as it does not satisfy the value


⇒ Option B is incorrect as it does not satisfy the value


⇒ Option D is incorrect as it does not satisfy the value


Question 34.

If and sec θ + cosec θ + x, then the value of x is
A. 1

B. 2

C.

D.


Answer:

Given, sinθ –cosθ = 0


⇒ squaring on both sides


⇒ (sin2θ + cos2θ -2sinθ cosθ) = 0


⇒ (1-2sinθ cosθ ) = 0



⇒ secθ + cosecθ = x




⇒ sinθ + cosθ = x sinθ cosθ


⇒ sinθ + cosθ =


Squaring on both sides




⇒ x2 = 8


⇒ x = 2√2


⇒ Option A is incorrect as it does not satisfy the value


⇒ Option B is incorrect as it does not satisfy the value


⇒ Option C is incorrect as it does not satisfy the value


Question 35.

If 2 cos 3 θ = 1, then the value of θ is
A. 10o

B. 15o

C. 20o

D. 30o


Answer:

Given,


⇒ we know


∴ θ = 20


⇒ cos3(20) = cos60


⇒ Option A is incorrect as it does not satisfy the value


⇒ Option B is incorrect as it does not satisfy the value


⇒ Option D is incorrect as it does not satisfy the value


Question 36.

Let us write whether the following statements are true or false :

i. If then the least value of (sec2 α + cos2 α) is 2.

ii. The value of is 1.


Answer:

(i) TRUE


⇒ let us consider α as 0°


⇒ sec20 + cos20 = 1 + 1 = 2


(ii) FALSE


⇒ cos90 = 0



Question 37.

Let us fill in the blanks :

i. The value of is ________.

ii. If then the value of cos θ is __________

iii. If then the value of cos4θ – sin4θ is __________


Answer:

(i) Given,



=


= 4


(ii) Given,


⇒ θ = 0


⇒ cos0° = 1


(iii) Given,


⇒ cos4θ –sin4θ = (cos2θ –sin2θ)(cos2θ + sin2θ)


=


=



Question 38.

If and then let us determine the values of both r and θ.


Answer:

Given, rcosθ = 2√3, rsinθ = 2





⇒ θ = 30





⇒ r = 4



Question 39.

If sin A + sin B = 2 where and then let us find out the value of (cos A + cos B).


Answer:

Given, sinA + sinB = 2


⇒ let A = B = 90°


⇒ sin90 + sin90 = 2


⇒ cosA + cosB = cos90° + cos90° = 0



Question 40.

If 0° < θ < 90°, then let us calculate the least value of (9 tan2 θ + 4 cot2 θ).


Answer:

Let θ = 45


⇒ (9tan2θ + 4cot2θ) [tan45 = 1, cot45 = 1]


⇒ 9 + 4


= 13



Question 41.

Let us calculate the value of (sin2 α + cos6 α + 3 sin2 α cos2 α).


Answer:

Given, sin6α + cos6α + 3sin2α cos2α


⇒ sin6α + cos6α + 3sin2α cos2α


= (sin2α)3 + (cos2α)3 + 3sin2α cos2α


= (sin2α + cos2α)(sin4α–sin2αcos2α + cos4α ) + 3sin2α cos2α


= sin4α + cos4α–sin2αcos2α + 3sin2α cos2α


= (sin2α + cos2α)2–2sin2αcos2α–sin2αcos2α + 3sin2α cos2α


= 1



Question 42.

If cosec2 θ = 2 cot θ and 0°<θ<90°, then let us determine the value of θ.


Answer:

Given, cosec2θ = 2cotθ


Let θ = 45


⇒ cosec2θ = (√2)2


= 2


⇒ 2cotθ = 2(1) = 2