If cos A + sin A = (√2) cos A, prove that cos A − sin A = (√2) sin A

 If cos A + sin A = (√2) cos A, prove that cos A − sin A = (√2) sin A

given

cos A + sin A = (√2) cos A

sin A =  (√2) cos A - cos A

sin A = (√2 - 1 ) cos A

sin A/ (√2 - 1 )  = cos A

rationalising

(√2 + 1 ) sin A/ [ (√2 - 1 ) (√2 + 1 ) ]  = cos A

(√2 + 1 ) sin A / [2-1] = cosA

(√2 + 1 ) sin A / [1] = cosA

(√2 + 1 ) sin A = cosA

apply distribution law

(√2) sin A + 1 sinA = cosA

(√2) sin A + sin A = cosA  

(√2) sin A  = cosA - sin A

OR

cos A − sin A = (√2) sin A

cbse 10th maths old board exam question paper 2025 2026 trigonometry

video of an alternate method 

Find ∫ [ √(tan x ) + √(cot x) ]dx step by step

 Find ∫ [ √(tan x ) + √(cot x) ]dx

more explanation in the video

Evaluate the indefinite integral

∫ [ √(tan x) + √(cot x) ] dx

Strategy overview
The most efficient route is to express tangent and cotangent through sine and cosine, then search for a substitution that treats the two square roots in a symmetric way. A natural attempt is to combine sine and cosine linearly because sin x and cos x appear in reciprocal fashion inside tan x and cot x. The expression sin x minus cos x is the key. With that choice you can leverage the identity for sin 2x and a compact derivative that collapses a messy sum into a familiar primitive.

Detailed solution
Start by rewriting the integrand through sine and cosine. Write tan x as sin x over cos x. Write cot x as cos x over sin x. Then the integrand becomes

√(sin x over cos x) + √(cos x over sin x)

This is

(√sin x over √cos x) + (√cos x over √sin x)

Now bring the two terms over a common style of denominator by noting that

(√sin x over √cos x) + (√cos x over √sin x)
equals
( sin x + cos x ) divided by √(sin x cos x)

To see that, write each fraction with denominator √(sin x cos x). The first term becomes sin x divided by √(sin x cos x). The second term becomes cos x divided by √(sin x cos x). Summing yields the expression above. The integrand is now

( sin x + cos x ) / √( sin x cos x )

This is a big step because it reduces the pair of nested roots to a single root. Next recall that sin 2x equals 2 sin x cos x. Thus sin x cos x equals sin 2x over 2. The denominator becomes √( sin 2x over 2 ). That is √(sin 2x) over √2. Therefore the integrand becomes

( sin x + cos x ) divided by ( √(sin 2x) over √2 )
which equals
√2 ( sin x + cos x ) / √( sin 2x )

You want to introduce a substitution whose derivative matches sin x plus cos x. Observe that the derivative of sin x minus cos x is cos x plus sin x. This is exactly the numerator. Let

t = sin x − cos x

Then dt = (cos x + sin x) dx. That means (sin x + cos x) dx equals dt. The integral becomes

∫ √2 dt / √( sin 2x )

We must now express sin 2x in terms of t. Compute t squared:

t² = (sin x − cos x)² = sin² x + cos² x − 2 sin x cos x = 1 − 2 sin x cos x

Hence 2 sin x cos x = 1 − t². But 2 sin x cos x equals sin 2x. Therefore sin 2x = 1 − t². The integrand is now completely in terms of t:

∫ √2 dt / √( 1 − t² )

This is the standard inverse trigonometric integral. The antiderivative is

√2 arcsin t + C

Now substitute back t = sin x − cos x to obtain the final expression

√2 arcsin( sin x − cos x ) + C

This form is accepted across board solutions and university notes because it displays the direct use of the template ∫ dt over √(1 − t²) equals arcsin t plus constant. Some manuals prefer to convert arcsin of a linear combination to an arctan form. That conversion is valid but not necessary for credit in most marking schemes. Present the arcsin form unless your class notes or teacher prefers a different inverse function. The equality of forms follows from standard relations among inverse trigonometric functions and from the identity that links t to sin 2x. For exam presentation keep the answer as shown with the constant of integration included.

Why this method fits widely used rubrics
Board mark schemes value clarity, a correct substitution, and a standard primitive. The steps above check all three boxes. First, the rewrite of the original integrand into a single fraction over √(sin x cos x) is transparent and short. Second, the substitution uses a linear combination of sine and cosine that appears often in these courses, so it looks familiar to graders. Third, the integral of dt over √(1 − t²) is a known template. If you write each of these moves cleanly, you match the expectations in CBSE Class 12 and ISC, in ICSE advanced sets, in Cambridge IGCSE Additional Mathematics extension exercises, in AS and A Level Pure Mathematics across CIE, Edexcel, OCR, and AQA, in IB Mathematics AA at both levels, and in AP Calculus BC. In JEE practice the same sequence gives a speed advantage because the derivative of sin x minus cos x produces the exact numerator, so you avoid detours.

Common slips and how to avoid them
The most frequent error is dropping the factor √2 when converting √( sin 2x over 2 ) to √(sin 2x) over √2. Keep that factor with care. Another slip is to forget the chain of identities that shows sin 2x equals 1 minus t². Build that bridge step by step using t² and the Pythagorean identity for sine and cosine. A third slip is to omit the constant of integration. Examiners look for it.

How to write the solution in a crisp mark scheme style
Begin with the conversion of the integrand to √2 times the fraction (sin x plus cos x) over √( sin 2x ). Introduce t as sin x minus cos x. Replace dt by the numerator times dx. Replace sin 2x by 1 minus t². Complete the integral to yield √2 arcsin t plus constant. Substitute back t. Close with the final expression. This sequence mirrors the solution skeleton provided in many board scoring rubrics and in university course keys.

Why the topic matters for learners moving between programs
Many students switch boards or programs during their senior years. A learner moving from a national board to IGCSE Additional Mathematics or to AS Level Pure will find that the underlying tools remain the same even though notation and pacing shift. The present integral is a friendly point of contact. It uses the identity for sin 2x. It uses a linear substitution. It uses one of the three classic inverse trigonometric primitives. That trio overlaps across curricula. As a result, mastering this single example builds confidence and transfers smoothly to adjacent problems that share the same structure, such as sums and differences of square roots built from sine and cosine ratios.

Connection to classroom assessment and computer based testing
Marking practices in school boards and in university courses place value on clean structure. You can score full credit by showing the three transitions and the final result without long commentary. Computer based practice systems also tend to accept any equivalent antiderivative. If your system presents a different inverse trigonometric function as the model answer, confirm equivalence by differentiation. Different forms that differ by a constant are equally valid. The arcsin form is compact and easy to differentiate, so it is a solid default in time bound conditions.

Tips for revision
Read the solution once without writing. Read it again and copy the key lines by hand. Then differentiate the final answer to check that you recover the original integrand. This last act links the algebra to the geometric meaning of inverse trigonometric functions and strengthens retention for examination day. If you study with a group, agree on the same structure so that peer feedback is consistent with rubric language.

Final answer to present in exams
∫ [ √(tan x) + √(cot x) ] dx = √2 arcsin( sin x − cos x ) + C

That is the complete result. It flows from a single substitution anchored in the identity for sin 2x. It meets the standards of clarity used in the boards and programs listed earlier, and it aligns with university calculus notes that stress method first, pattern recognition second, and presentation third. Place the video above this text in your BlogSpot post so that readers can watch the derivation and then scan the written solution for reinforcement. This pairing helps learners from different curricula follow a consistent set of steps. It also helps teachers and tutors share a uniform explanation that works across classroom settings and practice platforms.

2(sin(3pi/4))^2+2(cos(pi/4))^2 +2 (sec(pi/3))^2 = 10

ncert  cbse 11th trigonometry exercise 3.3

4. 2(sin(3pi/4))^2+2(cos(pi/4))^2 +2 (sec(pi/3))^2  = 10

using trigonometry formula trigonometry identities

sin(3pi/4)  = sin[pi - (pi/4)] = sin(pi/4) = 1 / [sqrt(2)]

cos(pi/4) =1 / [sqrt(2)]

sec(pi/3) = 2


LHS =

2(sin(3pi/4))^2+2(cos(pi/4))^2 +2 (sec(pi/3))^2 

= 2  {1 / [sqrt(2)] }^2 +2  {1 / [sqrt(2)] }^2 +2{2}^2

=2(1/2) +2(1/2) +2(4)

= 1 + 1 + 8 =10  =RHS

1. prove that [sin(pi/6)]^2+[cos(pi/3)]^2 -[tan(pi/4)]^2 = [-1/2]

using trigonometry formula trigonometry identities

sin(pi/6) =1/2

cos(pi/3) = 1/2

tan(pi/4) = 1

LHS =

[sin(pi/6)]^2+[cos(pi/3)]^2 -[tan(pi/4)]^2

=[1/2]^2 + [1/2]^2  - [1]^2

= [1/4] +[1/4] - 1

= [1/2] -1

=(-1/2) = RHS

3.3

1. prove that [sin(pi/6)]^2+[cos(pi/3)]^2 -[tan(pi/4)]^2 = [-1/2]
 solution

4. 2(sin(3pi/4))^2+2(cos(pi/4))^2 +2 (sec(pi/3))^2  = 10

solution

5.Find the values of sin 75degrees and tan 15degrees

solution

6. prove that cos[(pi/4)-x]cos[(pi/4)-y]- sin[(pi/4)-x]sin[(pi/4)-y] = sin(x+y)
 solution

7. [ tan{(pi/4)+x} ] /  [ tan{(pi/4 )- x} ]  = { [1+tanx] / [1-tanx] }^2
solution

8. prove that
[cos(pi+x)cos(-x)] / [sin(pi-x)cos((pi/2)+x) ] = [cotx]^2
 solution

9. prove that
cos[(3pi/2)+x]cos[2pi+x] {cot[(3pi/2)-x] + cot[2pi+x]} = 1
 solution

 
10. prove that sin[(n+1)x]sin[(n+2)x] +cos[(n+1)x]cos[(n+2)x] =cosx
 solution

11. prove that cos[(3pi/4)+x] - cos[(3pi/4)-x] = (-sqrt(2))sinx

solution

12.(sin6x)^2 - (sin4x)^2 = sin2x sin10x

solution

13.(cos2x)^2  - (cos6x)^2 = sin4x sin8x
 solution

14. Prove that sin2x+2sin4x+sin6x = 4[(cosx)^2]sin4x
solution

15.prove that cot4x[sin5x+sin3x]=cotx[sin5x-sin3x]
 solution

16.Prove that [cos9x -cos5x] / [sin17x - sin3x ] = -sin2x / cos10x
solution

17 prove that [sin5x + sin3x] / [cos5x+cos3x] = tan4x
solution

18. Prove that [sinx -siny] / [cosx +cosy] = tan[(x-y)/2]
solution

19.prove that [sinx + sin3x] / [cosx+cos3x] = tan2x
solution

20. Prove that [sinx - sin3x] / [ (sinx)^2 - (cosx)^2 ] = 2sinx
solution

21.Prove that [cos4x+cos3x+cos2x]/[sin4x+sin3x+sin2x] = cot3x
solution


22.Prove that cotx cot2x -cot2xcot3x-cot3xcotx = 1
solution

23. tan4x = { 4tanx{ 1 - [(tanx)^2] } } / { 1 - 6 [(tanx)^2] + [(tanx)^4]}
 solution


24. Prove that cos4x = 1-8[(sinx)^2][(cosx)^2]
solution 

25. Prove that cos6x = 32[cosx]^6 -48[cosx]^4 +18[cosx]^2 -1
 solution


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exercise 3.3 ncert trigonometry 12

exercise 3.3 ncert trigonometry 12

12.(sin6x)^2 - (sin4x)^2 = sin2x sin10x

using trigonometry formula trigonometry identities

sinx - siny =2cos[(x+y)/2] sin [(x-y)/2]

sinx + siny =2sin[(x+y)/2] cos[(x-y)/2]

finally use 2sinxcosx  = sin2x

Factorise and then apply the identities

LHS = {sin6x + sin4x } {sin6x - sin4x}

={2sin(10x/2)cos(2x/2)}{2cos(10x/2)sin(2x/2)}

={2sin5xcosx}{2cos5xsinx}

={ 2sinxcosx}{2sin5xcos5x} on regrouping

= {sin2x} {sin10x} using sin2x formula

=RHS

13.(cos2x)^2  - (cos6x)^2 = sin4x sin8x

using trigonometry formula trigonometry identities

cosx - cosy = -2sin[(x+y)/2] sin[(x-y)/2]

cosx + cosy =2cos[(x+y)/2] cos[(x-y)/2]

finally use 2sinxcosx  = sin2x

LHS =(cos2x)^2  - (cos6x)^2

={cos2x+cos6x} {cos2x-cos6x}

= { -2sin(8x/2)sin(-4x/2)} {2cos(8x/2)cos(-4x/2)}

= {-2sin4xsin(-2x)}{2cos4xcos(-2x)} {using sin(-x)= -sinx and cos(-x)=cosx}

={-2sin4x[-sin2x]}{2cos4xcos2x}

={2sin2xcos2x}{2sin4xcos4x} on regrouping

=sin4xsin8x

=RHS.

 

3.3

12.(sin6x)^2 - (sin4x)^2 = sin2x sin10x

solution

13.(cos2x)^2  - (cos6x)^2 = sin4x sin8x
 solution

17 prove that [sin5x + sin3x] / [cos5x+cos3x] = tan4x
solution

19.prove that [sinx + sin3x] / [cosx+cos3x] = tan2x
solution

20. Prove that [sinx - sin3x] / [ (sinx)^2 - (cosx)^2 ] = 2sinx
solution

21.Prove that [cos4x+cos3x+cos2x]/[sin4x+sin3x+sin2x] = cot3x
solution


22.Prove that cotx cot2x -cot2xcot3x-cot3xcotx = 1
solution

23. tan4x = { 4tanx{ 1 - [(tanx)^2] } } / { 1 - 6 [(tanx)^2] + [(tanx)^4]}
 solution


24. Prove that cos4x = 1-8[(sinx)^2][(cosx)^2]
solution 

25. Prove that cos6x = 32[cosx]^6 -48[cosx]^4 +18[cosx]^2 -1
 solution


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if A+B+C = 180° , show that cot A cotB + cotBcotC +cotC cotA=1

if A+B+C = 180° ,
show that

tanA +tanB + tanC = tanAtanBtanC and

cot A cotB + cotBcotC +cotC cotA=1



given A+B+C = 180°

implies A + B = 180°- C

implies tan(A + B) = tan(180°- C)

using trigonometry formulae

tan(A + B) = -tan(C)

therefore

[tanA +tanB] / [1-tanAtanB ] = -tanC

cross multiplying

tanA +tanB = -tanC[1-tanAtanB ]

tanA +tanB = -tanC +tanAtanBtanC

or

tanA +tanB + tanC = tanAtanBtanC

divide each term with tanAtanBtanC we get


cot A cotB + cotBcotC +cotC cotA=1

some other problems
If A+B+C=π,
prove that: sinA+sinB+sinC = cot(A/2).cot(B/2) [sinA+sinB-sinC]

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if (A-B) = pi/4 show that (1+tanA)(1+tanB) = 2tanA

if (A-B) = pi/4 show that (1+tanA)(1+tanB) = 2tanA

make it into A = B + (pi/4)
take tan on both sides, apply tan(A+B) identity
simplify and extract tanA
add tanA to both sides and then factorise
trigonometry identities

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collection of problems in integration, differentiation, areas, volumes, trigonometry, matrices, limits

Calculus
problems on integration ------------------------ integration

problems on application of integration like
area, volume, arc length etc ------------------------integration application

problems on differential equations ------differential equation

problems on differentiation------------------------ differentiation

problems on limits ------------------limits

problems on matrices --------------matrices

problems on trigonometry ------trigonometry


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conditional identities in trigonometry

conditional identities in trigonometry are usually based on A,B, C being angles of a triangle or A + B +C = π

examples

If A + B + C = π , show that cosA + cosB + cos C = 1 + 4 sin(A/2)sin(B/2)sin(C/2)

explanation of If A + B + C = π , show that cosA + cosB + cos C = 1 + 4 sin(A/2)sin(B/2)sin(C/2)

If A+B+C=π, prove that: sinA+sinB+sinC = cot(A/2).cot(B/2) [sinA+sinB-sinC]

explanation





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finding sin( arctan(2) )

find sin(arctan(2) )
let A = arctan(2) ------------------(1)
tanA = 2 = 2 / 1
use these on a triangle with side opposite to A as 2, adjacent to A as 1
use pythagoras theorem to get the hypotenuse sqrt(5)

so sin A = 2 / sqrt(5)

using equation(1)
sin ( arctan(2) ) = 2 / sqrt(5) = [ 2*sqrt(5) ] / 5



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proving a trigonometry identity

show that [( tanx -sinx ) / (2tanx) ] = sin² (x/2)
divide term by term and change
every thing in terms of sinx and cosx and
then use half angle formulae
use this link for trigonometric identities formulae
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trigonometric identities

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trigonometry problem

show that 1/(1-cosx) =(cscx+cotx)cscx

start from the LHS, introduce the conjugate
simplify the denominator using identity then divide term by term

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trigonometry formulae

for some other examples
prove that (cotA+cosecA-1)/(cotA-cosecA+1)=cosecA+cotA -----> (problem on trigonometry )

Prove that (tanxsinx) / (tanx+sinx) = (tanx-sinx) / (tanx sinx) -----> (problem on trigonometry )

find exact value of tan[22½ °] or tan (pi / 8 ) -----> (problem on trigonometry )

find exact value of tan15° without using calculator --------value of tan15° without using calculator

conditional identity : If A+B+C=π,
prove that: sinA+sinB+sinC = cot(A/2).cot(B/2) [sinA+sinB-sinC] -----> (problem on conditional identity (trigonometry) )


graphs of trigonometric functions-----> (graphs of trigonometric functions )

prove that
{ tan((π/4)+x) - tan((π/4)-x)} / { tan((π/4)+x) + tan((π/4)-x) } = sin2x ---------trigonometry problem

prove that (cos²x - sin²x) / (cos²x + sinxcosx) = 1 - tanx -------------trigonometry problem

show that
tan(x)/[1-cot(x)] + cot(x)/[1-tan(x)]= 1+sec(x)csc(x) , x≠ n π /4 ---------trigonometry


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to find cot(x ) if you are given that sin(x)=2/3 by drawing a triangle

find cot(x ) if you are given that sin(x)=2/3 and that x is acute.

draw a rt. triangle with one angle as x (not the right angle)



choose the opp. side as 2 and hypotenuse as 3 so that the third side is
sqrt(5) (using pythagoras theorem)

so cot(x) = (sqrt 5) /2


some trigonometry formulae

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