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Trigonometric Ratios

Class 10th Mathematics RS Aggarwal Solution
Exercise 5
  1. If sintegrate heta = root 3/2 , find all the values of all the trigonometric…
  2. If cosθ = 7/25, find all the values of all trigonometric ratios of θ.…
  3. If tanθ = 15/8, find all the values of all the trignometeric ratios of θ.…
  4. If cotθ = 2, find all the values of all the trigonometric ratios of θ.…
  5. If cosecθ = √10, find all the values of all the trignometeric ratios of θ.…
  6. If sintegrate heta = a^2 - b^2/a^2 + b^2 , find all the values of all the…
  7. If 15 cotA = 8, find the values of sinA and secA.
  8. If sinA = 9/41, find the values of cosA and tanA.
  9. If cosθ = 0.6, show that (5sinθ - 3tanθ) = 0
  10. If cosecθ = 2, show that cottheta + sintegrate heta /1+costheta = 2…
  11. If tanθ = 1/√7 , show that cosec^2theta -sec^2theta /cosec^2theta +sec^2theta =…
  12. If tanθ = 20/21, show that 1-sintegrate heta +costheta /1+sintegrate heta…
  13. If secθ = 5/4, show that sintegrate heta -2costheta /tantheta +cottheta = 12/7…
  14. If cotθ = 3/4, show that root sectheta -cosectheta /sectheta +cosectheta =…
  15. If sinθ = 3/4, show that root cosec^2theta -cot^2theta /sec^2theta -1 = root…
  16. If sinθ = a/b, show that sectheta +tantheta = root b+a/b-a
  17. If cosθ = 3/5, show that sintegrate heta -cottheta /2tantheta = 3/160…
  18. If tanθ = 4/3, show that sinθ + cosθ = 7/5
  19. If tanθ = a/b, show that asintegrate heta -bcostheta /asintegrate heta…
  20. If 3 tan = 4, show that
  21. If 3cotθ = 2, show that 4sintegrate heta -3costheta /2sintegrate heta…
  22. If 3cotθ = 4, show that 1-tan^2theta /1+tan^2theta = cos^2theta -sin^2theta…
  23. If secθ = 17/8, verify that 3-4sin^2theta /4cos^2theta -3 = 3-tan^2theta…
  24. In the adjoining figure, ∠B = 90°, ∠ BAC = θ°, BC = CD = 4cm, AD = 10cm. find…
  25. In a Δ ABC, ∠ B = 90°, AB = 24cm and BC = 7cm. Find a) sinA b) cosA c) sinC d)…
  26. In a Δ ABC, ∠ C = 90°, ∠ ABC = θ° , BC = 21units and AB = 29units. Show that…
  27. In a Δ ABC, ∠ B = 90°, AB = 12cm and BC = 5cm. Find i. cosA ii. cosecA iii.…
  28. If sinα = 1/2, prove that 3cosα - 4cos^3 α = 0
  29. In a Δ ABC, ∠ B = 90° and tanA = 1/√3. Prove that i. sinA cosC + cosA sinC = 1…
  30. If ∠ A and ∠ B are acute angles such that sinA = sinB then prove that ∠ A = ∠…
  31. If ∠ A and∠ B are acute angles such that tanA = tanB then prove that ∠ A = ∠ B.…
  32. In a right ΔABC, right - angled at B, it tanA = 1 then verify that 2sinA cosA =…
  33. In the given figure of ΔPQR, ∠ P = θ° and ∠ R = ϕ°. Find a. root x+1 cotphi b.…
  34. If x = cosecA + cosA and y = cosecA - cosA the prove that (2/x+y)^2 + (x-y/2)^2…
  35. If x = cotA + cosA and y = cotA - cosA, prove that (x-y/2)^2 + (x-y/x+y)^2 = 1…
Exercise 5
Question 1.

If , find all the values of all the trigonometric ratios of θ.
Answer:

We have,

Let Perpendicular = √3 k
and hypotenuse = 2k
where 'k' is some integer.



By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴AB2 = BC2 + AC2


(2k)2 = (k√3)2 + AC2


4k2 = 3k2 + AC2


AC2 = (4 - 3)k2


AC2 = k2


→ AC = k, for some number k


Hence, the trigonometric ratios for the given θ are:


sinθ =


cosθ = AC/AB = k/(2k) = 1/2


tanθ = BC/AC = sinθ /cosθ = (k√3)/k = √3


cotθ = 1/tanθ = AC/BC = k/(k√3) = 1/√3


cosecθ = 1/sinθ = AB/BC = (2k)/(k√3) = 2/√3


secθ = 1/cosθ = AB/AC = (2k)/k = 2
Question 2.

If cosθ = 7/25, find all the values of all trigonometric ratios of θ.
Answer:

We have, cosθ = (7k)/(25k) = base/hypotenuse (For some value of k)



By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴AB2 = BC2 + AC2


= (25k)2 = BC2 + (7k)2 (for some value of k)


= 625k2 = BC2 + 49k2


= BC2 = 576k2


= BC2 = (24k)2


→ BC = 24k


Hence, the trigonometric ratios of the given θ are:


sinθ = BC/AB = (24k)/(25k) = 24/25


cosθ = 7/25


tanθ = BC/AC = sinθ /cosθ = 24/7


cotθ = AC/BC = 1/tanθ = 7/24


cosecθ = AB/BC = 1/sinθ = 25/24


secθ = AB/AC = 1/cosθ = 25/7
Question 3.

If tanθ = 15/8, find all the values of all the trignometeric ratios of θ.
Answer:

We have, tanθ = 15k/8k = perpendicular/base (For some value of k)



By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴AB2 = BC2 + AC2


AB2 = (15k)2 + (8k)2


AB2 = 225k2 + 64k2


AB2 = 289k2 = (17k)2


→ AB = 17k


Hence, the trignometeric ratios for the given θ are:


sinθ = BC/AB = (15k)/(17k) = 15/17


cosθ = AC/AB = (8k)/(17k) = 8/17


tanθ = 15/8


cotθ = AC/BC = 1/tanθ = 8/15


cosecθ = AB/BC = 1/sinθ = 17/15


secθ = AB/AC = 1/cosθ = 17/8
Question 4.

If cotθ = 2, find all the values of all the trigonometric ratios of θ.
Answer:

We have, cotθ = 2k/k = base/perpendicular (For some value of k)



By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴AB2 = BC2 + AC2


AB2 = (k)2 + (2k)2


AB2 = k2 + 4k2


AB2 = 5k2 = (k√5)2


→ AB = k√5


Hence, the trignometeric ratios for the given θ are:


sinθ = BC/AB = k/(k√5) = 1/√5


cosθ = AC/AB = (2k)/(k√5) = 2/√5


tanθ = BC/AC = sinθ /cosθ = k/(2k) = 1/2


cotθ = 2


cosecθ = AB/BC = 1/sinθ = √5


secθ = AB/AC = 1/cosθ = √5/2
Question 5.

If cosecθ = √10, find all the values of all the trignometeric ratios of θ.
Answer:

We have, cosecθ = (k√10)/k = 1/sinθ (For some value of k)


→ sinθ = k/(k√10) = perpendicular/hypotenuse



By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴AB2 = BC2 + AC2


(√10k)2 = (k)2 + AC2


AC2 = 10k2 - k2


AC2 = 9k2 = (3k)2


→ AC = 3k


Hence, the trignometeric ratios for the given θ are:


sinθ = 1/√10


cosθ = AC/AB = (3k)/(k√10) = 3/√10


tanθ = BC/AC = sinθ /cosθ = 1/3


cotθ = AC/BC = 1/tanθ = 3


cosecθ = √10


secθ = AB/AC = 1/cosθ = √10/3
Question 6.

If , find all the values of all the trignometeric ratios of θ.
Answer:

We have, sinθ = = perpendicular/hypotenuse (For some value of k)



By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴AB2 = BC2 + AC2


{(a2 + b2)k}2 = {(a2 - b2)k}2 + AC2


a4k2 + b4k2 + 2a2b2k2 = a4k2 + b4k2 - 2a2b2k2 + AC2


AC2 = 4a2b2k2 = (2abk)2


→ AC = 2abk


Hence, the trigonometric ratios for the given θ are:


sinθ =


cosθ = AC/AB = =


tanθ = BC/AC = sinθ /cosθ =


cotθ = AC/BC = 1/tanθ =


cosecθ = AB/BC = 1/sinθ =


secθ = AB/AC = 1/cosθ =



Question 7.

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

We have, 15 cotA = 8


→ cotA = (8k)/(15k) = 1/tanA = AC/BC (For some value of k)



By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴AB2 = BC2 + AC2


AB2 = (15k)2 + (8k)2


AB2 = 225k2 + 64k2


AB2 = 289k2


= (17k)2


→ AB = 17k


∴ sinA = BC/AB = (15k)/(17k) = 15/17


secA = 1/cosA = AB/AC = (17k)/(8k) = 17/8
Question 8.

If sinA = 9/41, find the values of cosA and tanA.
Answer:

We have, sinA = (9k)/(41k) = BC/AB (For some value of k)



By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴AB2 = BC2 + AC2


= (41k)2 = (9k)2 + AC2


= 1681k2 = 81k2 + AC2


AC2 = 1600k2


= (40k)2


→ AC = 40k


∴ cosA = AC/AB = (40k)/(41k) = 40/41


tanA = BC/AC = sinA/cosA = 9/40
Question 9.

If cosθ = 0.6, show that (5sinθ - 3tanθ ) = 0
Answer:

We have cosθ = 0.6 = (6k)/(10k) = AC/AB (For some value of k)



By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴AB2 = BC2 + AC2


= (10k)2 = BC2 + (6k)2


= 100k2 = BC2 + 36k2


= BC2 = 64k2


= (8k)2


→ BC = 8k


∴ sinθ = BC/AB = (8k)/(10k) = 0.8


tanθ = sinθ /cosθ = 0.8/0.6


consider, the LHS,


5sinθ - 3tanθ = 5(0.8) - 3(0.8/0.6)


= 4 - 3(0.4/0.3)


= 4(0.3) - 3(0.4)


= 1.2 - 1.2


= 0


= RHS


HENCE PROVED
Question 10.

If cosecθ = 2, show that


Answer:

We have, cosecθ = 2 = 1/sinθ


⇒ θ = 30°


consider LHS
= cotθ +
= √3 +

= √ 3 +

=

=

=

= 2

= RHS


HENCE PROVED
Question 11.

If tanθ = 1/√7 , show that


Answer:

We have, tanθ = k/(k√7) = BC/AC (For some value of k)



By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴AB2 = BC2 + AC2


= AB2 = (k)2 + (k√7)2


= AB2 = k2 + 7k2


= 8k2 = (2k√2)2


→ AB = 2k√2


∴ cosecθ = AB/BC = 2k√2


secθ = AB/AC = =


consider the LHS,


LHS =


=


=


= 48/64


= 3/4


= RHS


HENCE PROVED
Question 12.

If tanθ = 20/21, show that


Answer:

We have, tanθ = (20k)/(21k) = BC/AC (For some value of k)



By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴AB2 = BC2 + AC2


= AB2 = (20k)2 + (21k)2


= AB2 = 400k2 + 441k2


= AB2 = 841k2


= (29k)2


→ AB = 29k


∴ sinθ = BC/AB = (20k)/(29k) = 20/29


cosθ = AC/AB = (21k)/(29k) = 21/29


consider, the LHS


LHS = =


=


= 30/70


= 3/7


= RHS


HENCE PROVED
Question 13.

If secθ = 5/4, show that


Answer:

We have, secθ = 5/4 = 1/cosθ


→ cosθ = (4k)/(5k) = AC/AB (For some value of k)



By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴AB2 = BC2 + AC2


= (5k)2 = BC2 + (4k)2


= 25k2 = BC2 + 16k2


= BC2 = 9k2


→ BC = 3k


∴ sinθ = BC/AB = (3k)/(5k) = 3/5


consider the LHS


LHS = =


=


=


=


=


= 12/7


= RHS


HENCE PROVED
Question 14.

If cotθ = 3/4, show that


Answer:

We have, cotθ = (3k)/(4k) = AC/BC (For some value of k)



By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴AB2 = BC2 + AC2


= AB2 = (4k)2 + (3k)2


= AB2 = 16k2 + 9k2


= AB2 = 25k2


= (5k)2


→ AB = 5k


∴ sinθ = BC/AB = (4k)/(5k) = 4/5


cosθ = AC/AB = (3k)/(5k) = 3/5


consider the LHS


LHS =


=


=


=


= 1/√7


= RHS


HENCE PROVED
Question 15.

If sinθ = 3/4, show that


Answer:

We have, sinθ = (3k)/(4k) = BC/AB (For some value of k)


Consider the LHS


LHS = =


=


=


= cosθ /sinθ


= cotθ



By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴AB2 = BC2 + AC2


= (4k)2 = (3k)2 + AC2


= 16k2 = 9k2 + AC2


= AC2 = 7k2


→ AC = k√7


∴ cotθ = AC/BC = k/(k√7) = 1/√7


∴ LHS = RHS


HENCE PROVED
Question 16.

If sinθ = a/b, show that


Answer:

Consider LHS,


LHS = secθ + tanθ =


= (1)


We have, sinθ = (ak)/(bk) = BC/AB (For some value of k)



By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴AB2 = BC2 + AC2


= (bk)2 = (ak)2 + AC2


= AC2 = b2k2 - a2k2


→ AC =


∴ cosθ = AC/AB = =


∴ from(1)


LHS = =


=


=


= RHS


HENCE PROVED
Question 17.

If cosθ = 3/5, show that


Answer:

We have, cosθ = (3k)/(5k) = AC/AB (For some value of k)



By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴AB2 = BC2 + AC2


= (5k)2 = BC2 + (3k)2


= 25k2 = BC2 + 9k2


= BC2 = 16k2


= (4k)2


→ BC = 4k


∴ sinθ = BC/AB = (4k)/(5k) = 4/5


tanθ = BC/AC = sinθ /cosθ = (4k)/(3k) = 4/3


cotθ = 1/tanθ = 3/4


consider the LHS


LHS = =


=


= 3/160


= RHS


HENCE PROVED
Question 18.

If tanθ = 4/3, show that sinθ + cosθ = 7/5
Answer:

We have, tanθ = (4k)/(3k) = BC/AC (For some value of k)



By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴AB2 = BC2 + AC2


= AB2 = (4k)2 + (3k)2


= AB2 = 16k2 + 9k2


= AB2 = 25k2


= (5k)2


→ AB = 5k


sinθ = BC/AB = (4k)/(5k) = 4/5


cosθ = AC/AB = (3k)/(5k) = 3/5


consider LHS = sinθ + cosθ =


= 7/5


= RHS


HENCE PROVED
Question 19.

If tanθ = a/b, show that


Answer:

Consider the LHS =


(Dividing the numerator and denominator by cos θ)


=


=


=


= RHS


HENCE PROVED
Question 20.

If 3 tan θ = 4, show that


Answer:

We have, 3 tan θ = 4


→ tan θ = 4/3 (1)


Consider the LHS =


(Dividing the numerator and denominator by cosθ )


=


= (from(1))


=


= 8/10


= 4/5


= RHS


HENCE PROVED
Question 21.

If 3cotθ = 2, show that


Answer:

We have 3cotθ = 2


→ cotθ = 2/3 (1)


Consider the LHS =


(Dividing the numerator and denominator by sinθ )


=


=


= (from(1))


=


= 2/6


= 1/3


= RHS


HENCE PROVED
Question 22.

If 3cotθ = 4, show that


Answer:

Consider the LHS =


=


=


= cos2θ – sin2θ (∵ cos2θ + sin2θ = 1)


= RHS


HENCE PROVED
Question 23.

If secθ = 17/8, verify that


Answer:

We have, secθ = (17k)/(8k) = AB/AC (For some value of k)



By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴AB2 = BC2 + AC2


= (17k)2 = BC2 + (8k)2


= 289k2 = BC2 + 64k2


= BC2 = 225k2


→ BC = 15k


∴ sinθ = BC/AB = (15k)/(17k) = 15/17


cosθ = AC/AB = (8k)/(17k) = 8/17


tanθ = BC/AC = sinθ /cosθ = 15/8


consider the LHS =


=


=


=


= ( - 33)/( - 611)


= 33/611


Now consider RHS =


=


=


= ( - 33)/( - 611)


= 33/611


∴ RHS = LHS


HENCE PROVED
Question 24.

In the adjoining figure, ∠B = 90°, ∠ BAC = θ°, BC = CD = 4cm, AD = 10cm. find

i. sinθ

ii. cosθ
Answer:

Clearly, Δ ABC and Δ ABD are right angled triangles


where AD = 10cm BC = CD = 4cm


BD = BC + CD = 8cm



Applying Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴AD2 = BD2 + AB2


= (10)2 = (8)2 + AB2


= 100 = 64 + AB2


= AB2 = 36


= (6)2


→ AB = 6cm


Now applying Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴AC2 = BC2 + AB2


= AC2 = (4)2 + (6)2


= AC2 = 16 + 36


= 52


→ AC = √52


= 2√13cm


i. sinθ = BC/AC =


= 2/√13


= (2√13)/13


ii. cosθ = AB/AC =


= 3/√13


= (3√13)/13
Question 25.

In a Δ ABC, ∠ B = 90°, AB = 24cm and BC = 7cm. Find

a) sinA

b) cosA

c) sinC

d) cosC
Answer:

Clearly Δ ABC is a right angled triangle,



By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴ AC2 = BC2 + AB2


= AC2 = (7)2 + (24)2


= AC2 = 49 + 576


= AC2 = 625


→ AC = 25


a) sinA = BC/AC = 7/25


b) cosA = AB/AC = 24/25


c) sinC = AB/AC = 24/25


d) cosC = BC/AC = 7/25
Question 26.

In a Δ ABC, ∠ C = 90°, ∠ ABC = θ° , BC = 21units and AB = 29units. Show that cos2θ –sin2θ = 41/841
Answer:

Δ ABC is a right angled triangle



By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴AB2 = AC2 + BC2


= (29)2 = AC2 + (21)2


= 841 = AC2 + 441


= AC2 = 400


→ AC = 20


∴ sinθ = AC/AB = 20/29


cosθ = BC/AB = 21/29


cos2θ –sin2θ = (21/29)2 - (20/29)2


=


= 41/841


= RHS


HENCE PROVED
Question 27.

In a Δ ABC, ∠ B = 90°, AB = 12cm and BC = 5cm. Find

i. cosA

ii. cosecA

iii. cosC

iv. cosecC
Answer:

Δ ABC is a right angled triangle



By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴AC2 = BC2 + AB2


= AC2 = (5)2 + (12)2


= AC2 = 25 + 144


= AC2 = 169


= (13)2


→ AC = 13


i. cosA = AB/AC = 12/13


ii. cosecA = AC/BC = 13/5


iii. cosC = BC/AC = 5/13


iv. cosecC = AC/AB = 13/12
Question 28.

If sinα = 1/2, prove that 3cosα - 4cos3α = 0
Answer:

We have, sinα = k/(2k) = BC/AB (For some value of k)



By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴AB2 = BC2 + AC2


= (2k)2 = (k)2 + AC2


= 4k2 = k2 + AC2


= AC2 = 3k2


→ AC = k√3


∴ cosα = AC/AB = (k√3)/(2k) = √3/2


Then, 3cosα - 4cos3α =


=


= 0


LHS = RHS


HENCE PROVED
Question 29.

In a Δ ABC, ∠ B = 90° and tanA = 1/√3. Prove that

i. sinA cosC + cosA sinC = 1

ii. cosA cosC - sinA sinC = 0
Answer:

We have, tanA = k/(k√3) = BC/AB


Δ ABC is a right angled triangle



By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴AC2 = BC2 + AB2


= AC2 = (k)2 + (k√3)2


= AC2 = k2 + 3k2


= AC2 = 4k2


→ AC = 2k


∴ sinA = BC/AC = k/(2k) = 1/2


cosA = AB/AC = (k√3)/(2k) = √3/2


sinC = AB/AC = (k√3)/(2k) = √3/2


cosC = = BC/AC = k/(2k) = 1/2


i. sinA cosC + cosA sinC = (1/2)(1/2) + ( √3/2)( √3/2)


=


= 4/4


= 1


∴ RHS = LHS


HENCE PROVED


ii. cosA cosC - sinA sinC = (1/2)(√3/2) - (1/2)(√3/2)


= (√3/4) - (√3/4)


= 0


∴ RHS = LHS


HENCE PROVED
Question 30.

If ∠ A and ∠ B are acute angles such that sinA = sinB then prove that ∠ A = ∠ B.
Answer:

Consider ΔABC to be a right - angled triangle.



∴ sinA = BC/AB


sinB = AC/AB


Given that sinA = sinB


BC/AB = AC/AB


→ BC = AC


→ ∠ A = ∠ B (In a triangle, angles opposite to equal angles are equal)
Question 31.

If ∠ A and∠ B are acute angles such that tanA = tanB then prove that ∠ A = ∠ B.
Answer:

Consider ΔABC to be a right - angled triangle.



∴ tanA = BC/AC


tanB = AC/BC


Given that tanA = tanB


BC/AC = AC/BC


BC2 = AC2


→ BC = AC


→ ∠ A = ∠ B (In a triangle, angles opposite to equal angles are equal)
Question 32.

In a right ΔABC, right - angled at B, it tanA = 1 then verify that 2sinA cosA = 1
Answer:

Consider ΔABC to be a right - angled triangle.



tanA = 1 = BC/AB


→ BC = AB


Also, tanA = 1 = sinA/cosA


→ sinA = cosA


By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴AC2 = BC2 + AB2


= AC2 = 2BC2


= (AC/BC)2 = 2


= AC/BC = √2


→ cosecA = √2


→ sinA = 1/√2 = cosA


2 sinA cosA = 2(1/√2)( 1/√2)


= 2(1/2)


= 1


∴ LHS = RHS


HENCE PROVED
Question 33.

In the given figure of ΔPQR, ∠ P = θ° and ∠ R = ϕ°. Find

a.

b.

c. Cosθ
Answer:

Δ PQR is a right - angled triangle


By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2


∴PR2 = RQ2 + PQ2


= (x + 2)2 = x2 + PQ2


= x2 + 4 + 4x = x2 + PQ2


PQ2 = 4 + 4x


→ PQ = 2


a. cotϕ = RQ/PQ =


∴ × =


= x/2


b. tanθ = RQ/PQ =


∴ =


= x×


= x2/2


c. cosθ = PQ/RP =



Question 34.

If x = cosecA + cosA and y = cosecA - cosA the prove that


Answer:

x + y = cosecA + cosA + cosecA - cosA


= 2cosecA


x - y = cosecA + cosA - (cosecA - cosA)


= cosecA + cosA - cosecA + cosA


= 2cosA


Consider the LHS, =


= sin2A + cos2A - 1


= 1 - 1 (∵sin2A + cos2A = 1)


= 0


= RHS


HENCE PROVED
Question 35.

If x = cotA + cosA and y = cotA - cosA, prove that


Answer:

x - y = cotA + cosA - (cotA - cosA)


x - y = cotA + cosA - cotA + cosA


x - y = 2cosA


x + y = cotA + cosA + cotA - cosA


x + y = 2cotA


Consider the LHS =


= cos2A + sin2A


= 1 (∵sin2A + cos2A = 1)


= RHS


HENCE PROVED