## 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

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)

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. rivika manya says:

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. rivika manya says:

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).

• Hi, The answer is cos 2A = cos^2 A – sin^2 A = 2cos^2 A -1 = 1-2sin^ 2 A

(Sudheer)

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

• SUBHAM SEKHAR PATRA says:

VERY EASY…
TAKE TAN AND COT IN TERMS OF SIN AND COS..
U WIL GET UR ANS.

• Subham sekhar patra says:

its very easy..
just take tan and cot as sin/cos and cos/sin..and go on simplifying..
u will get RHS

8. R. Guragain says:

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

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

• Srinivas says:

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

11. reshmi says:

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

Hi Reshmi,

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,

12. tanisha says:

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

Hi Tanisha,

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,

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….

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,

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) }

• Saurav Anand says:

sorry, i posted wrong question , it has some mistakes.

16. Saurav Anand says:

I want answer of the following question.
Prove that

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

Hi Saurav,
Hope it helps,
Cheers,

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

Hi Akhtar,

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
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.

• Subbu says:

No da..it’s good n difficult!!