Class 10–Proving trigonometric identities

Trigonometry being a branch of mathematics, which deals with the ratios of right angled triangles.  Trigonometric identities are equalities that involve trigonometric functions and are true for every single value of the occurring variables.

Proving trigonometric identities using Pythagoras

Right triangle and trigonometric identities

The above triangle is a right angled triangle. Let us prove trigonometric identities by using Pythagoras theorem.

Pythagoras theorem states that AC2 = AB2 + BC2.

Therefore, dividing each term of the theorem by AC2, we get:

(AC2/ AC2) = (AB2/AC2) + (BC2/AC2)

i.e. (AB/AC)2 + (BC / AC)2 = (AC / AC)2.

i.e. (cos A)2 + (sin A)2 = 1

Therefore, cos2 A + sin2 A = 1

This is true for all A such that 00 ≤ A ≤ 900. Therefore, this is a trigonometric identity.

Now, dividing each term by (AB)2, we get :

AC2/AB2 = AB2 / AB2 +  BC2/AB2

i.e. (AB/AB)2 + (BC / AB)2 = (AC / AB)2.

Therefore, 1 + tan2 A = sec2 A

This is also true for all A such that 00 ≤ A ≤ 900. Therefore, this is also a trigonometric identity.

Now, dividing each term by (BC)2, we get :

i.e. (AB/BC)2 + (BC / BC)2 = (AC / BC)2.

Therefore, cot2 A + 1 = cosec2 A

This is also a trigonometric identity.

Therefore there are 3 trigonometric identities. They are

  1. cos2 A + sin2 A = 1
  2. 1 + tan2 A = sec2 A
  3. cot2 A + 1 = cosec2 A

Problems on trigonometric identities

Let us see some problems on these identities. The problems will be based on the above three trigonometric identities. Also knowledge on trigonometric ratios helps in solving these problems.

1) Solve (sec A + tan A) (1 – sin A)

Solution:     (sec A + tan A) (1 – sin A) = sec A + tan A – sin A sec A – sin A tan A

= (1/cos A) + (sin A/cos A) – sin A (1/cos A) – sin A (sin A/cos A)

= (1 + sin A – sin A – sin2 A)/ cos A

= (1- sin2 A)/ cos A                                       [we know that cos2 A + sin2 A = 1]

= cos2 A/ cos A

= cos A

2) Prove (1+ sec A)/ sec A = sin2A / (1 – cos A)

Solution:  Let us take the RHS of the equation

sin2A / (1 – cos A) = (1 – cos2A)/(1 – cos A)     [we know that cos2 A + sin2 A = 1]

= (1 + cos A)(1 – cos A) / (1 – cos A)

= (1 + cos A)

Multiply and divide by sec A

= (1 + cos A) sec A/sec A

= (sec A + 1) / sec A

Hence it is proved.

3) Prove (cos A – sin A + 1)/ (cos A + sin A – 1) = cosec A + cot A.

Solution:    Let us take the left hand side of the equation. Divide numerator and denominator by sin A.

(cos A – sin A + 1)/ (cos A + sin A – 1) = (cot A – 1 + cosec A)/ (cot A + 1 – cosec A)

= [cot A – (1 – cosec A)]/[cot A + (1-cosec A)]

Multiply numerator and denominator by cot A – (1 – cosec A)

= [cot A – (1 – cosec A)]2/ [cot2A – (1-cosec A)2]

= [cot2A + (1- cosec A)2 – 2 cot A(1- cosec A)]/ [cot2 A – 1 – cosec2 A + 2 cosec A ]

= [cot2A + 1 + cosec2 A - 2 cosec A – 2 cot A + 2 cot A cosec A]/[2 cosec A – 2]

= [cosec2A + cosec2 A - 2 cosec A – 2 cot A + 2 cot A cosec A]/[2 cosec A – 2]

= [2cosec2 A - 2 cosec A – 2 cot A + 2 cot A cosec A]/[2 cosec A – 2]

= [2 cosec A(cosec A – 1) + 2 cot A(cosec A – 1)]/[2 cosec A – 2]

= (2 cosec A + 2 cot A)(cosec A – 1)/2(cosec A – 1)

= 2(cosec A + cot A)/2

= cosec A + cot A

Hence it is proved.

Practice problems on trigonometric identities

Solve the problems given below using trigonometric identities. Answers are also provided for these problems. The practice problems can be solved, once you understood the example problems given above.

1)      Solve  1/(cot2B + 1)   +   1/(tan2B + 1)

2)      Find tan2Bcos2B + 1 – sin2B

3)      Solve (tanB sinB) / (sec2B – 1)

Answers:

1)      1

2)      1

3)     Cos B

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53 Responses to Class 10–Proving trigonometric identities

  1. sakshi singh says:

    nice. got interesting sums from different people to solve. love trigonometry.

  2. hiii…….
    pls solve my problems…
    (cosec A – sin A) (sec A – cos A) = 1/tan A + cot A

    • A) LHS :

      (cosec A – sin A)(sec A – cos A) = (1/sin A – sin A) (1/cos A – cos A)
      (since cosec A = 1/sin A and sec A = 1/cos A)

      =(1-sin2A)/sinA * (1-cos2A)/cosA
      (since sin2A + cos2A = 1)

      = sin2A/cosA

      RHS:

      1/(tan A + cot A) = 1/(sin A/cos A + cos A/sin A)

      = 1/[(sin2A + cos2A)/sinAcosA]
      (since sin2A + cos2A = 1)

      = 1/(1/sinAcosA)

      = sinAcosA

      LHS = RHS

      Hence proved.

      (Sudheer Kumar)

  3. its urgent please solve this equation of trigonometry………

    1/cosec A – cot A – 1/sin A = 1/sin A – 1/cosec A + 1/cot A

    • (From Sudheer Kumar)
      Let us take the LHS of the equation 1/(cosecA-cotA)-1/sinA
      Multiply and divide 1/(cosecA – cotA) by (cosecA+cotA)
      1/(cosecA-cotA)-1/sinA = (cosecA+cotA) / (cosecA-cotA)(cosecA+cotA) – 1/sinA

      = (cosecA+cotA) / (cosec2A – cot2A) – cosecA

      = (cosecA+cotA) – cosecA
      [since cosec2A – cot2A = 1]

      = cotA

      Let us take the RHS of the equation 1/sinA-1/(cosecA+cotA)

      Multiply and divide 1/(cosecA + cotA) by (cosecA-cotA)

      1/sinA-1/(cosecA+cotA) = 1/sinA – (cosecA-cotA) / (cosecA-cotA)(cosecA+cotA)
      = cosecA – (cosecA-cotA) / (cosec2A – cot2A)
      = cosecA – (cosecA-cotA) [since cosec2A – cot2A = 1]
      = cotA
      LHS = RHS.
      Hence proved.

  4. farah says:

    hi. who can help me with this? please?
    If sin 2A = 2sin A cos A, cos 2A = cos2 A – sin2 A, show in 4 steps that tan 2A = (2 tan A) / (1 – tan2 A).

  5. Tamizh selvan says:

    Kindly send the solution for this problem

    If tanA =ntanB and sinA =msinB, then prove that cos2A = m2-1/n2-1

    • A) sinA = msinB
      sinB = sinA/m ….(1)
      tanA = n tanB
      sinA/cosA = n sinB/cosB
      sinA/sinB = ncosA/cosB
      m = ncosA/cosB
      cosB = n/m cosA …(2)
      (1)2+(2)2 = sin2B + cos2B = (sinA/m)2 + (n/m cosA)2
      => 1 = sin2A/m2 + n2cos2A/m2
      => m2 = (1-cos2A) + n2cos2A
      => m2 -1 = +n2cos2A-cos2A
      => (m2 -1) = (n2-1)cos2A
      => cos2A = (m2 -1)/ (n2-1)
      Hence proved.

      (Sudheer Kumar)

  6. It is best tution centre wher i learn any subject…and there are also very nice questins thnx…..

  7. SAJALA says:

    PLEASE GIVE A SOLUTION TO FOLLOWING PROBLEM. TanA/1-CotA+CotA/1-TanA= 1+ SecA CosecA

  8. R. Guragain says:

    I need the solution of cos pi by 7 + cos 2pi by 7 + cos 3pi by 7 = 1/8

    • padmaja says:

      Hi there,
      Please note that the question is incorrect and given below is the question and the answer.

      Q) cos pi/7 cos 2pi/7 cos 3pi/7 = 1/8
      A) Let us take LHS .
      Multiply & divide the equation with 2sin pi/7
      (2sin pi/7) ( cos pi/7 cos 2pi/7 cos 3pi/7) / (2sin pi/7)
      = (2sin pi/7 cos pi/7) (cos 2pi/7 cos 3pi/7) / (2 sin pi/7)
      = (sin 2pi/7) (cos 2pi/7 cos 3pi/7)/(2 sin pi/7)
      Multiply & divide the equation with 2
      = (2 sin 2pi/7 cos 2pi/7 cos 3pi/7)/(4 sin pi/7)
      = (sin 4pi/7 cos 3pi/7) / (4 sin pi/7)
      Cos 3pi/7 = -cos(pi – 3pi/7) = -cos 4pi/7
      So, (sin 4pi/7 cos 3pi/7) / (4 sin pi/7) = -(sin 4pi/7 cos 4pi/7)/(4 sin pi/7)
      Multiply & divide the equation with 2
      = -(2 sin 4pi/7 cos 4pi/7)/(8 sin pi/7)
      = (-sin 8pi/7) / (8 sin pi/7)
      sin 8pi/7 = sin(pi+pi/7) = -sin pi/7
      So, (-sin 8pi/7) / (8 sin pi/7) = (sin pi/7)/(8 sin pi/7)
      = 1/8
      = RHS
      Hence proved.

  9. krishnakali jana says:

    thanks

  10. Srinivas says:

    Urgent…. plz solve the problem.

    show that (cos^2A – 3 cosA + 2)/sin^A = 1

  11. reshmi says:

    find 1+cosA/1-cos A=tan^A/(secA-1)^2

    • padmaja says:

      Hi Reshmi,
      Please find the answer to your question.

      LHS:
      Step 1: Multiply numerator & denominator by (1-cosA)
      Step 2: (1+cosA)(1-cosA)/(1-cosA)(1-cosA) = (1-cos2A)/(1-cosA)2
      = sin2A/(1-cosA)2
      Step 3: Divide numerator & denominator by cos2A & you will get the RHS i.e. tan2A/(secA-1)2

      Cheers,
      Padmaja

  12. tanisha says:

    i want the ans 2 dis question very urgently………please help me

    (1+tan x +cot x)(sin x-cos x)=sec x/cosec sq.x-cosec x /sec sq. x

    • padmaja says:

      Hi Tanisha,
      Please find the answer to your problem.

      LHS:
      (1 + tanA + cotA)(sinA – cosA) = sinA((1 + tanA + cotA) – cosA(1 + tanA + cotA)
      = sinA + sinA tanA + sinA cotA – cosA – cosA tanA – cosA cotA
      = sinA + sinA sinA/cosA + sinA cosA/sinA – cosA – cosA sinA/cosA – cosA cosA/sinA
      = sinA + sin2A/cosA + cosA – cosA – sinA – cos2A /sinA
      = sin2A/cosA – cos2A /sinA
      = sin2A secA – cos2A cosecA
      = secA/cosec2A – cosecA/sec 2A
      = RHS
      Hence proved

      Cheers,
      Padmaja

  13. Mayank tuli says:

    Hiiiiiiii
    My name is Mayank tuli of std 10. Sir i study my NCERT book i also try too solved another books . But in our NCERT text book their is a chapter INTRODUCTION OF TRIGONOMETRY. and in another books their is only chapter TRIGONOMETRY. THEIR is an another question which is not in our book how wesolved their problem. If their is an another book u please told too me….

    • padmaja says:

      Hi Mayank,
      Trigonometry chapter introduces you to the subject and also covers the various ways by which to solve a problem. Using your knowledge and the explanation provided in the textbook, try to solve them yourself.
      Cheers,
      Padmaja

  14. Saurav Anand says:

    I want answer of the following question.

    Prove that:-)

    (1 + tanA + cotA)(sinA – cosA) = (secA/cosec sq. A – cosecA/sec sq. A)

  15. Saurav Anand says:

    I Want the answer of the following question.

    Prove that

    Cot2A(SecA-1) / (1+SinA) = Sec2A { (1-SinA)/(1+SecA) }

  16. Saurav Anand says:

    I want answer of the following question.
    Prove that

    sinA/(secA+tanA-1) + cosA/(cosecA+cotA-1) = 1

  17. Akhtar Hussain says:

    I also want to make member of your sight.
    I want to send proves of trigonometric identities

  18. Akhtar Hussain says:

    Gave a effective tips for teaching of trigonometry.

  19. Akhtar Hussain says:

    Thank you so much to solve the problem.
    i am from pakistan tehsil Zafarwal
    my profession is teaching.

  20. Akhtar Hussain says:

    solve the problem sin2A+cos2A is equa to 1 why give its logic

    • padmaja says:

      Hi Akhtar,
      Here’s the answer to your question.

      Consider a right angled triangle with length of opposite side as x and length of adjacent side as y.
      Therefore length of hypotenuse = sqrt(x2+y2).
      Note: Here opposite side is the side that is opposite to angle A and adjacent side is the side that is adjacent to angle A.
      We know that sinA = opposite side/hypotenuse
      cosA = adjacent side/hypotenuse
      sinA = x/sqrt(x2+y2) = > sin2A = x2/(x2+y2)
      cosA = y/sqrt(x2+y2) => cos2A = y2/(x2+y2)
      Therefore sin2A + cos2A = x2/(x2+y2) + y2/(x2+y2)
      = (x2+y2)/ (x2+y2)
      = 1.
      Hence proved.

  21. Akhtar Hussain says:

    I like so much this edurites official blog

  22. PIRASANNAH says:

    Hai.I want the solution for this question.
    cot(a)+tan(2A) = cot(A)sec(2A)

    plzzzzzz urgent……..

  23. Sujaya says:

    Need A Solution
    If TanA= (TanB+TanC)/(1+TanATanB)
    Prove That:
    Sin2A=(Sin2B+Sin2C)/(1+Sin2BSin2C)

  24. Praveen says:

    Hi,I just want answer for the following question.
    Prove That.
    1/cosecA-cotA-1/sinA=1/sinA-1/cosecA+cotA

  25. Praveen says:

    Hi,I just want answer for the following question.
    Prove That.
    1/cosecA-cotA-1/sinA=1/sinA-1/cosecA+cotA

    • abhijeet says:

      1 – cos^2 A = sin^2 A
      => (1 – cosA) * (1 + cosA) = sinA * sinA
      => (1 + cosA) / sinA = sinA / (1 – cosA)
      => (1 + cosA – sinA) / (sinA – 1 + cosA) = (1 + cosA) / sinA = 1/sinA + cosA/sinA

      … [because a/b = c/d => (a-c)/(b-d) = a/b]

      => (cosA – sinA + 1) / (cosA + sinA – 1) = cosecA + cotA

  26. anamika says:

    i want the solution of
    Prove that – sin A/cot A +cosec A=2+ sin A/cot A-cosec A

  27. mm says:

    hi i want a solution for a problem tanA/SecA-1 – sinA/1+cosA=2cotA
    WANT IT URGENTLY..

  28. nandu says:

    what is tan60+sin40*cosec45/sec35
    if u dont answer then i will tell u

    try hard
    best of luck!!!!!!!!!!!!!!!!!!!!!!!!!

  29. rajesh1100 says:

    please find out my problem cos sq A- SIN SQ A = TAN SQ B GIVEN PROVE THAT 2 COS SQB-1=COS SQ A-SIN SQB =TEN SQA

  30. yash says:

    very very easy questions u have posted.

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