Trigonometry is yet another scoring section of CGL. Most of the questions can be solved with

Just like Algebra, where we assumed the values of variables, in trigonometry we will assume the value of 'theta'. And like algebra, make sure the value you are assuming for theta will not make the denominator zero.

1. When you don't have to deal with fractions you can assume ϴ = 90 or 0

E.g. (acosϴ - bsinϴ)

2. When fraction is given and putting ϴ = 90 or 0 is making denominator zero, then you can go with ϴ = 45.

3. Don't assume a value for theta at which the trigonometric function is not defined. E.g. When tanϴ is given, you can't assume ϴ = 90.

4. When you are assuming two angles, go with A = 60 and B = 30

Note : These are not hard and fast rules and you can assume any value you like, but make sure

You only need to memorize the values of sin, cos and tan (for ϴ = 0, 30, 45 and 60). The values of cosec, sec and cot can be obtained by reciprocating sin, cos and tan respectively.

Now let's solve CGL questions

Put ϴ = 90

*jugaad*.Just like Algebra, where we assumed the values of variables, in trigonometry we will assume the value of 'theta'. And like algebra, make sure the value you are assuming for theta will not make the denominator zero.

**How to assume value -**1. When you don't have to deal with fractions you can assume ϴ = 90 or 0

E.g. (acosϴ - bsinϴ)

2. When fraction is given and putting ϴ = 90 or 0 is making denominator zero, then you can go with ϴ = 45.

3. Don't assume a value for theta at which the trigonometric function is not defined. E.g. When tanϴ is given, you can't assume ϴ = 90.

4. When you are assuming two angles, go with A = 60 and B = 30

Note : These are not hard and fast rules and you can assume any value you like, but make sure

**denominator**≠ 0. Sometimes when you assume ϴ, you may end up with two options(say A and B) that are giving similar results (but two options will still get eliminated, i.e., C and D). Now change the value of theta and check only A and B.You only need to memorize the values of sin, cos and tan (for ϴ = 0, 30, 45 and 60). The values of cosec, sec and cot can be obtained by reciprocating sin, cos and tan respectively.

Now let's solve CGL questions

x = a(1 + 0) = a

y = b(1 - 0) = b

Again assume ϴ = 90

Then a*1 + b*0 = cor a = c

You have to find the value of acosϴ - bsinϴ

acosϴ - bsinϴ = -b (Since ϴ = 90)

Now put a = c in all the 4 options to check which one can give '-b' as the output

Assume ϴ = 45

(tanA - secA - 1)/(tanA + secA + 1) = -√2/(2 + √2) = -1/(√2 + 1) = 1 - √2 (rationalize)

Put ϴ = 45 in all the 4 options and check which one will give (1 - √2) as the output

A) √2 - 1

B) √2 + 1

D) √2 - 1

Sometimes you would need to assume two angles

Let A = 60 and B = 30

Then n = 3 and m = √3

cos

Now put n = 3 and m = √3 in all the options and check which one will give 1/4

you can write, secA - tanA = 1/p ... (2)

Now add (1) and (2)

2secA = p + 1/p

In the above question, p=2

So secA = 5/4

Now we have to find sinA. The best way to determine the value of a trigonometric function when the value of other function is given, is by making a triangle.

secϴ = Hypotenuse/Base

Here secA = 5/4, hence hypotenuse = 5 and base = 4, which means perpendicular = 3

sinA = perpendicular/hypotenuse = 3/5 = 0.6

So x

^{2}/a^{2}+ y^{2}/b^{2}= 2**Answer : C**

Some complex questions can be solved by assuming the value of theta

Then a*1 + b*0 = cor a = c

You have to find the value of acosϴ - bsinϴ

acosϴ - bsinϴ = -b (Since ϴ = 90)

Now put a = c in all the 4 options to check which one can give '-b' as the output

**Answer : D**

Assume ϴ = 45

a (tan45 + cot45) = 1

So a = 1/2

sin45 + cos45 = b

So b = √2

Put a = 1/2 and b = √2 in all the options and check which among the 4 equations is right (i.e. LHS should be equal to RHS)

**Answer : A (both LHS and RHS are equal to 1)**

(tanA - secA - 1)/(tanA + secA + 1) = -√2/(2 + √2) = -1/(√2 + 1) = 1 - √2 (rationalize)

Put ϴ = 45 in all the 4 options and check which one will give (1 - √2) as the output

A) √2 - 1

B) √2 + 1

**C) 1 - √2**D) √2 - 1

**Answer : C**Sometimes you would need to assume two angles

Let A = 60 and B = 30

Then n = 3 and m = √3

cos

^{2}A = 1/4 (since A = 60)Now put n = 3 and m = √3 in all the options and check which one will give 1/4

**Answer : B**

**There is one type of question which is frequently asked by SSC -**

**When you see secA + tanA = something (let's say 'p') ... (1)**

you can write, secA - tanA = 1/p ... (2)

Now add (1) and (2)

2secA = p + 1/p

**secA = (p**^{2}+ 1)/2p [You can memorize this formula]**tanA = (p**^{2}- 1)/2pIn the above question, p=2

So secA = 5/4

Now we have to find sinA. The best way to determine the value of a trigonometric function when the value of other function is given, is by making a triangle.

secϴ = Hypotenuse/Base

Here secA = 5/4, hence hypotenuse = 5 and base = 4, which means perpendicular = 3

sinA = perpendicular/hypotenuse = 3/5 = 0.6

**Answer : C**

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Pls upload more stuff so tht ve can practice more ....thnkew :)

ReplyDeletePls upload more stuff so tht ve can practice more ....thnkew :)

ReplyDeleteIf you liked the blog, please don't forget to share it with others. After all sharing is caring :)

DeleteSure :)

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ReplyDeleteDo you wear a cape or do you think its just outdated now for superheroes 😁

ReplyDeleteThank you Prashant, for sharing short tricks to solve cubersome questions. Please share similar tricks, if any, for

ReplyDeleteWork-Time, STD and geometry.

Super stuff, please share some stuff related to trigonometry questions on maximum and minimum values. Please its a bit confusing, all the efforts are really appreciated😊

ReplyDeleteJust landed on ur blogspot. Now not want to leave it soon. Its Fantastic,.......... Thank u sir for these wonderful tricks.

ReplyDeletewhat is the formula for finding perpendicular with hypotenuse and base - pls answer

ReplyDeleteHypotenuse^2 = Perpendicular^2 + Base^2

Deletethank you forgot pythogoras for a while

DeleteThanks sir .

ReplyDeletePlease upload more tricks

If SinA +cosA = √2cosA ,then find value of cotA

ReplyDeletedivide LHS and RHS with sinA...

DeleteDivide LHS and RHS by Cos A

DeleteThis gives

TanA = √2-1

CotA = √2+1

Sir, facing difficulty in solving some questions in trigonometry. Posting those questions. Will be of thankful if you can kindly give some short tricks to solve them ,

ReplyDelete1. if cotA + cosecA = 3, and A is an acute angle, find value of cosA

2. if (a^2-b^2) sinα + 2ab cosα = a^2+b^2 , then tanα = ?

3. if α is positive acute angle and 2sinα + 15cos2α = 7 , then cotα=?

4. the least value of (4sec^2α + 9cosec^2α) = ?

5. if sinα + cosα = √2 cos(90-α) then cotα = ?

1.Multiply LHS and RHS with sinA. Put cosA = x and sinA = root(1 - x^2). Solve the quadratic

Delete2. Put a=b, cosA = 1, so A = 0, tanA = 0

3. Put cos2A = 1 - 2(sinA)^2. Now whole equation is in the form of sinA. Solve the quadratic to get the value of sinA. From sinA, calculate cotA

4. = 4tan^2 + 4 + 9cot^2 + 9

= 13 + 4tan^2 + 9cot^2 ... (1)

Now we have to find minimum value of 4tan^2 + 9cot^2

AM ≥ GM

(4tan^2 + 9cot^2)/2 ≥ √(4tan^2.9cot^2)

(4tan^2 + 9cot^2) ≥ 12

So minimum value of (4tan^2 + 9cot^2) is 12

Put this value in (1)

Answer : 25

5. cos(90-A)=sinA

So cosA = sinA(√2-1)

cotA = √2-1

Thank you soo much sirfor ur time .

ReplyDeleteSir, no. 1 ka logic Put cosA = x and sinA = root(1 - x^2)., ye wala part samajh nahi aaya , why are we putting the value sinA = root(1 - x^2) ? and sir, AM aur GM ka matlab kya hai ?

For your 1st doubt -

Deletesin^2 + cos^2 = 1 (thats how)

for 2nd doubt-

AM = arithmetic mean

AM of two numbers a and b is (a+b)/2

GM = geometric mean

GM of two numbers a and b is root(a.b)

AM ≥ GM is a fact

Got it sir . Thanks .. :)

ReplyDeleteHello Sir,

ReplyDeleteAny trick to solve this question,

[Cotx/(cotx-cot3x)+tanx/(tanx-tan3x)] is equal to:-

a) 0 b) 1 c) -1 d) 2

Conventional method is time taking.

Looking forward for your response.:)

Put x=15, and

Deletetan15 = 2-√3 (Mug these values)

cot15 = 2+√3

Answer : 1

if we put x=45 then we will get the same answer

DeleteThank you Sir for quick reply.feeling happy..:)

ReplyDeleteOne more question,

If (a^2-b^2)sinx + 2abcosx = a^2+b^2,then the value of tanx is:

1) (a^2+b^2)/2ab 2) (a^2-b^2)/2 3)(a^2-b^2)/2ab 4)(a^2+b^2)/2

This question is again time taking.

put theta=45

Deleteu'll get a^2(1-root 2)-b^2(1+root2)+2abc=0

put a=1,b=1,c=root2

now in option put a=1 b=1 and check which option will yeild 1 as answer as tan4 is

Sir what should i do to master trigo questions?

ReplyDeleteDo i need to learn all trigo formulae from my 11th and 12th standard maths books?

sir , not able to solve some trigonometry questions .

ReplyDelete1. Find the value of 3 (sinx - cosx)^4 + 6 (sinx + cosx)^2 + 4 (sin^6x + cos^6x)

a)14, b)11 , c) 12, d) 13

2.if xcosA - sinA =1 , then x^2 +(1+x^2) =?

a)2 b)1 , c) -1 d) 0

3. if cosX + cos^2X =1, then sin^8x + 2sin^6x + sin^4x =?

a)0 b)3 , c) 2 d) 1

Please guide sir ..

1. put x= 90

ReplyDelete2. put a=0

3. a^2+b^2+2ab = (a+b)^2

Thank you sir..

Delete1. put x= 90

ReplyDelete2. put a=0

3. a^2+b^2+2ab = (a+b)^2

how to solve below problem in fast way

ReplyDelete(secx + tan x)\(sec x - tan x) = 209/79

please tell me if there is any trick to solve this in min time

multiply numerator and denominator with cosx. Then cross multiply to find sinx

Deletethanks a lot sir.

Deleteplease give some english hacks if there are any to score more in english

Sir in q no 1 we can also take theta=zero?

ReplyDeleteYes, take what ever value you like, but make sure denominator is not equal to zero. In ques 1, you can take theta = 45 as well

DeleteGr8 Work Sir

ReplyDeletePrashant, MEA is my dream job...plz post something exiting abt MEA that will boost my confidence..

ReplyDeleteU r doing a great job indeed..

Simply awesome bro...

ReplyDeleteSuperb sir

ReplyDeleteThank u

& definitely

I will recommend this blog to many ssc cgl aspirants

Sir, in this..

ReplyDeleteSin^2A+Sin^2B = 2..

Then Cos(A+B)/2 = ?

If I put A=30 and B=60 in 1st equation I would get answer as 1.. But RHS is 2.. How would I proceed...?

you only have to assume such values that will satisfy the equation...put A=B=90

DeleteHence cos(A+B)/2 = cos90 = 0

Ok, sir.. Thanks..

DeleteThis comment has been removed by the author.

ReplyDeleteSir the trick used in question number 5 i.e SEC A=p^2 +1/2p can be valid for COSEC A as well as TAN A=p^2-1/2p can be valid for COT A?? I've tried solving and by this and the answer for COT A and COSEC A was coming right. Please verify it once and thank you sir for your hacks, they are really helpful.

ReplyDeleteSir, from which identity did we deduce the fact that

ReplyDeleteif secA+tanA=p ,

then, secA-tanA=1/p

Sir in question no.2, if we put theta equals to 0 then we are getting option c.... Then how to conclude answer of this question?

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Sec theta = x+1/4x

ReplyDeleteFind sec theta + tan theta