- Algebra: 3 formulas
- Trigonometry: Formulas
- Case: Only trig formulas
- Case: Combo of Trig + algebra formulas
- Mock Questions
Before moving to the next topic of trigonometry, I would like to quote the tweets of SSC chairman’s official account:
What SSC chief said
- In SSC CGL 2013, the Maths will be of class 10th level.(tweet link)
- …questions in trigonometry and geometry will certainly be easier.(tweet link)
- (because)… some non mathematics candidates (in 12th std) had complained about trigonometry and geometry questions of Higher Secondary level (class12) level (in earlier SSC CGL exam).(tweet link)
Point being: there is no need to be afraid of either trigonometry or geometry. You need to understand a few concepts, mugup a few formulas and practice questions, then both trigonometry and geometry can be solved without much problem. Anyways, back to the trig. Topics. Until now we saw
- Height and distance problems
- How to construct the trigonometry table and solve questions
- Complimentary angles (90-A)
- now moving to the last major topic: the questions based on combo of Trig+algebra formulas. Almost all of them can be solved by mugging up only six formulas: 3 from algebra and 3 from trigonometry.
Algebra: 3 formulas
These three formulas are
- (A plus or minus B)2
- A2-B2
- (A plus or minus B)3
Let’s see
First formula
Positive sign | Negative sign |
(a+b)2=a2+2ab+b2 | (a-b)2=a2-2ab+b2 |
Second formula
(a2-b2)=(a+b)(a-b)
Third formula
Positive sign | Negative |
a3+b3=(a+b)3-3ab(a+b) | a3-b3=(a-b)3+3ab(a-b) |
based on above formulas, we can derive some more formulas for example
(a+b)2=a2+2ab+b2=> a2+b2=(a+b)2-2ab
Same way,
a3+b3=(a+b)3-3ab(a+b)
=> (a+b)3= a3+b3+3ab(a+b)= a3+b3+3ab2+3a2b
Trigonometry: Formulas
Again, only three formulas
- Sin2a+cos2a=1
- Sec2a-tan2a=1
- Cosec2a-cot2a=1
you can verify these formulas by plugging the values from our trigonometry table.
It is important not to make mistake in these three fomulas vs complimentary angle
Complimentary angle (90-A) | Trig.formulas |
Sin and Cos | Here also both are together. Sin and cos: Sin2a+cos2a=1 |
Cosec and sec | Here cosec comes with cot (and not Sec). Cosec2a-cot2a=1 |
Tan and cot | Here Tan comes with sec (and not Cot): Sec2a-tan2a=1 |
Based on these formula, you can derive many other formulas for example
Tan=sin/cos.
And we now know that Sin2a+cos2a=1. So
cos2A=1-sin2a
cosA=sq.root of (1-sin2a)
now pluggin this back into Tan’s ratio
TanA=sinA/cosA
=sinA/ sq.root(1-sin2A)
In some books you might have seen such complex formulas for cosec, sec, cot also. Basically they’re derived from the three main formulas given above..
Case: Only trig formulas
Q. Find the value of sin2a+[1/(1+tan2a)]
We know the Sec2a-tan2a=1=>sec2a=1+tan2a
Substituting this in the question
sin2a+1/(1+tan2a)
=sin2a+(1/sec2a) ; but cos= 1/sec
= Sin2a+cos2a
=1
Q. if 3cos2A+7sin2A=4 then find value of cotA, given that A is an acute angle?
Approach
Instead of 7sin2a=3sin2a+4sin2a (because 4+3=7)….eq1
Now let’s look at the question
3cos2a+7sin2a=4
3cos2a+3sin2a+4sin2a =4 ; using value from eq1
3(cos2a+sin2a) +4sin2a =4
3(1)+ 4sin2a =4
4sin2a=4-3
Sin2A=1/4
sinA=squareroot (1/4)=1/2
From the trigonometry table (or Topi Triangles), we know that sin30=1/2 therefore angle “A” has to be 30 degrees. And from the same trigonometry table, we can see that cot30=root3. That’s our final answer.
Q. if cos2a+cos4a=1, what is the value of tan2a+tan4a
Approach
We know that Sin2a+cos2a=1 =>sin2a=1-cos2a…eq1
In the question, we are given that
cos2a+cos4a=1
taking cos2a on the right hand side
cos4a=1-cos2a
taking value from eq1
cos4a=sin2a
cos2a x cos2a = sin2a ; laws of surds and indices
cos2a=(sin2a/cos2a)
cos2a=tan2a……eq2
now lets move to the next part. in the question, we have to find the value of
tan2a+tan4a
=tan2a+(tan2a)2 ; because (a2)2=a2×2=4
= cos2a+cos4a ;applying value from eq2
=1 ;this was already given in the first part of the question itself.
Final answer=1.
Case: Combo of Trig + algebra formulas
Q.find value of sin4a-cos4a
We know that
(a2-b2)=(a+b)(a-b)
So instead of sin4a-cos4a, I can write
(sin2a)2-(cos2a)2 ; because a4=2×2 =(a2)2
=(Sin2a+cos2a)(Sin2a-cos2a)
=1 x (Sin2a-cos2a) ; because (a2-b2)=(a+b)(a-b)
=(Sin2a-cos2a)
We can further simplify this as (sina-cosa)x(sina-cosb)
Q. if sin4a-cos4a=-2/3 then what is the value of 2cos2a-1
Approach
We know that Sin2a+cos2a=1=> sin2a=1- cos2a….(eq1)
From the previous sum, we already know that
LHS | RHS |
sin4a-cos4a= | (sin2a-cos2a) |
(1-cos2a-cos2a) ; applying eq.1 | |
-2/3 (given in the Qs.) | =1-2cos2a |
2/3 | 2cos2a-1 ; multiplying both sides with (-1) |
Therefore, 2cos2a-1=2/3
Q. what is the value of (cosecA-sinA)2+(secA-cosA)2-(cotA-tanA)2
I’m dividing this equation into three parts. After that, I’ll apply the formula (a-b)2=a2-2ab+b2 on each of them
(coseca-sina)2 | +(seca-cosa)2 | -(cota-tana)2 |
Cosec2a-2cosecA x sinA+sin2a | +sec2a-2secA x cosA+cos2a | -(cot2A-2cotA x tanA +tan2A) |
If you observe the bold parts: 2 cosec A x sin A= 2 x 1 (because cosec and sin are inverse of each other so sin x cosec =1). Same situation will repeat two other parts (because sec x cos = 1 and tan x cot =1.)
So now our three parts will look like:
Cosec2a-2+sin2a | +sec2a-2+cos2a | -(cot2A-2 +tan2A) |
=Cosec2a-2+sin2a+ sec2a-2+cos2a- cot2A+2 -tan2A ;donot make silly mistakes while opening brackets with minus sign e.g. -(a+b-c)=-a-b+c.
Now let’s club them according to the three trig. Formulas (e.g.sin2a+cos2a=1)
= (Cosec2a- cot2A) +(sin2a+ cos2a)+( sec2a -tan2A)+(-2-2+2)
=1+1+1-2 ;because each of those pair is equal to 1.
=1
Q. If sinA-cosA=7/13, what is the value of sinA+cosA, given that A is an acute angle?
we know that
(a-b)2=a2-2ab+b2
LHS | RHS |
(sinA-cosA)2= | Sin2a-2sinA x cosA +cos2A |
(7/13)2= | 1-2sinA x cosA; because Sin2A+cos2A=1 |
Therefore, 2sinA x cosA=1-(7/13)2=….eq1
Moving forward, we also know that
(a+b)2=a2+2ab+b2
LHS | RHS |
(sinA+cosA)2= | Sin2a+2sinA x cosA +cos2A |
1+2sinA x cosA; because Sin2A+cos2A=1 | |
1+1-(7/13)2 ; plugging the value from eq.1 | |
(sinA+cosA)2= | =289/169=(17/13)2 |
Taking square-root on both sides,
sinA+cosA=17/13, that’s our final answer.
Mock Questions
- Find value of Cos2a+(1/1+cot2a)
- 0
- 1
- 2
- 3
- Find value of 2sin2a+4sec2a+5cot2a+2cos2a-4tan2a-5cosec2a
- 0
- 1
- 2
- 3
- Find the value of (cosA-sinA)2+(cosA+sinA)2
- 0
- 1
- 2
- 3
- Find the value of Cot2A x (sec2A-1)
- 0
- 1
- 2
- 3
- Find the value of (secA x cotA)2 – (cosec A x cosA)2
- 0
- 1
- 2
- 3
- secA/(cotA+tanA) will be equal to
- cosA
- cosecA
- sinA
- tanA
- (1+tanA+secA)(1+cotA-cosecA) will be equal to
- 0
- 1
- 2
- 3
- Cos6A+sin6A=
- 1-3(cosA x sinA)2
- 1+3(cosA x sinA)2
- 1-3(cosA x sinA)3
- 1-3(cosA x sinA)6
- (sin2A x cos2B) – (cos2A x sin2B) will be equal to
- Sin2A-cos2A
- Sin2A+cos2A
- Cos2A – cos2A
- Sin2A-Sin2B
- (tanA+cotA)(secA-cosA)(cosecA-sinA)=
- 0
- 1
- 2
- 3
- If Tan2A+Tan4A=1 then what is the value of cos2A+cos4A, given that A is an acute angle?
- 0
- 1
- 2
- 3
- (cosecA-cotA)2=
- (1-cosA)/(2+cosA)
- (1+cosA)/(1-cosA)
- (1-cosA)/(1+cosA)
- (1-cosA)x(1+cosA)
- [(tanA+secA)2-1]/[(tanA+secA)2+1], will be equal to
- tanA
- secA
- sinA
- cosA
- [(sin220+sin270)/(sec250-cos240)]+2cosec258-2(cot58xtan32)-(4tan13 x tan37 x tan45 x tan53 x tan77)
- 0
- -1
- 1
- None of above
- If cotA=root7, what is the value of (cosec2A-sec2A)/( cosec2A+sec2A)
- 3/2
- 3/4
- 4/3
- 2/3
- (sinA+cosecA)2+(cosA+secA)2 will be equal to
- Tan2A+cot2A+4
- Tan2A-cot2A+5
- Tan2A-cot2A
- Tan2A+cot2A+7
Answers
1)b,2)b, 3)c, 4)b, 5)b, 6)c , 7)c, 8)a, 9)d, 10)b, 11)b, 12)c,13)d, 14)b, 15)b,16)d
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Please someone explain problem 15..!
cotA= √7 that implies tanA=1/√7
cot2A = 7 and tan2A =1/7
now consider the equation
(cosec2A-sec2A)/( cosec2A+sec2A)
=((1+cot2A)-(1+tan2A))/ ((1+cot2A)+(1+tan2A))
=(1+cot2A-1- tan2A)/ (1+cot2A+1+ tan2A)
=(cot2A- tan2A)/ (2+cot2A+ tan2A)
=(7-1/7)/(2+7+1/7)
=(48/7)/(64/7)
=48/64
=3/4
If cos¢+cos€ +cos¥ =0
Then find sin^1028¢+sin^1028€ +sin^1028¥= ?
Sir need drdo, isro question and answer and short cut trick for all