Trigonometry is yet another scoring section of CGL. Most of the questions can be solved with 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
Put ϴ = 90
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
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
cos2A = 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 = (p2 + 1)/2p [You can memorize this formula]
tanA = (p2 - 1)/2p
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
Answer : C
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So x2/a2
+ y2/b2 = 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
cos2A = 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 = (p2 + 1)/2p [You can memorize this formula]
tanA = (p2 - 1)/2p
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
Answer : C
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Buy SSC Hack Book






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 :)
DeleteSuper sir
ReplyDeleteSuper sir
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?
ReplyDeleteSec theta = x+1/4x
ReplyDeleteFind sec theta + tan theta
SecA+tanA/secA-tanA=209/79 then find the value of sinA
ReplyDelete