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 AC^{2} = AB^{2} + BC^{2}.

Therefore, dividing each term of the theorem by AC^{2}, we get:

(AC^{2}/ AC^{2}) = (AB^{2}/AC^{2})^{ }+ (BC^{2}/AC^{2})

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

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

Therefore, **cos ^{2} A + sin^{2} A = 1**

This is true for all A such that 0^{0} ≤ A ≤ 90^{0}. Therefore, this is a trigonometric identity.

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

AC^{2}/AB^{2} = AB^{2 }/ AB^{2} + BC^{2}/AB^{2}

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

Therefore, **1 + tan ^{2} A = sec^{2} A**

This is also true for all A such that 0^{0} ≤ A ≤ 90^{0}. 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, **cot ^{2} A + 1 = cosec^{2} A**

This is also a trigonometric identity.

Therefore there are 3 trigonometric identities. They are

**cos**^{2}A + sin^{2}A = 1**1 + tan**^{2}A = sec^{2}A**cot**^{2}A + 1 = cosec^{2}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 – sin^{2} A)/ cos A

= (1- sin^{2} A)/ cos A [we know that **cos ^{2} A + sin^{2} A = 1]**

** **= cos^{2} A/ cos A** **

= **cos A**

**2) ****Prove (1+ sec A)/ sec A = sin ^{2}A / (1 – cos A)**

Solution: Let us take the RHS of the equation

**sin ^{2}A / (1 – cos A) **= (1 – cos

^{2}A)/(1 – cos A) [we know that

**cos**

^{2}A + sin^{2}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}/ [cot^{2}A – (1-cosec A)^{2}]

= [cot^{2}A + (1- cosec A)^{2} – 2 cot A(1- cosec A)]/ [cot^{2} A – 1 – cosec^{2} A + 2 cosec A ]

= [**cot ^{2}A + 1** + cosec

^{2}A - 2 cosec A – 2 cot A + 2 cot A cosec A]/[2 cosec A – 2]

= [cosec^{2}A + cosec^{2} A - 2 cosec A – 2 cot A + 2 cot A cosec A]/[2 cosec A – 2]

= [2cosec^{2} 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/(cot^{2}B + 1) + 1/(tan^{2}B + 1)

2) Find tan^{2}Bcos^{2}B + 1 – sin^{2}B

3) Solve (tanB sinB) / (sec^{2}B – 1)

**Answers:**

1) 1

2) 1

3) Cos B

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

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)

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.

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)

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)

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

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

VERY EASY…

TAKE TAN AND COT IN TERMS OF SIN AND COS..

U WIL GET UR ANS.

its very easy..

just take tan and cot as sin/cos and cos/sin..and go on simplifying..

u will get RHS

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.

thanks

Urgent…. plz solve the problem.

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

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

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

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

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

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

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,

Padmaja

I want answer of the following question.

Prove that:-)

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

I Want the answer of the following question.

Prove that

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

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

I want answer of the following question.

Prove that

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

Hi Saurav,

Please find the answer to your question by clicking on the link given below. Trigonometry

Hope it helps,

Cheers,

Padmaja

I also want to make member of your sight.

I want to send proves of trigonometric identities

Gave a effective tips for teaching of trigonometry.

Thank you so much to solve the problem.

i am from pakistan tehsil Zafarwal

my profession is teaching.

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

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.

I like so much this edurites official blog

Hai.I want the solution for this question.

cot(a)+tan(2A) = cot(A)sec(2A)

plzzzzzz urgent……..

Need A Solution

If TanA= (TanB+TanC)/(1+TanATanB)

Prove That:

Sin2A=(Sin2B+Sin2C)/(1+Sin2BSin2C)

Hi,I just want answer for the following question.

Prove That.

1/cosecA-cotA-1/sinA=1/sinA-1/cosecA+cotA

Hi Praveen,

The answer to your question can be found by clicking on the following link. Hope it helps.

http://www.edurite.com/answer/Trigonometry/2425

All the best,

Padmaja

Hi,I just want answer for the following question.

Prove That.

1/cosecA-cotA-1/sinA=1/sinA-1/cosecA+cotA

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

i want the solution of

Prove that – sin A/cot A +cosec A=2+ sin A/cot A-cosec A

Check the link given below for solution to your problem.

Pls solve without interchanging the side.

http://www.edurite.com/answer/Trigonometric%20identities/1584

pls solve without interchanging the side

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

WANT IT URGENTLY..

Hi Maitreyee,

You will find the solution to your problem in the following link. Please go through it.

http://www.edurite.com/answer/Trigonometric%20identities/1321

Hope this helps.

Regards,

Padmaja

what is tan60+sin40*cosec45/sec35

if u dont answer then i will tell u

try hard

best of luck!!!!!!!!!!!!!!!!!!!!!!!!!

Morning Nandu,

The answer to your question can be seen in the link given below. Hope it solves your query.

http://www.edurite.com/answer/Trigonometric%20identities/623

Regards,

Padmaja

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

Hi Rajesh,

You can find the solution to the given problem when you click on the given link below.

http://www.edurite.com/answer/trigonometry/535.

very very easy questions u have posted.

No da..it’s good n difficult!!

Hi Subbu,

Nothing is difficult. Just a little effort from your side will make it only good and not difficult.