- 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}
- A^{2}-B^{2}
- (A plus or minus B)^{3}
Let’s see
First formula
Positive sign | Negative sign |
(a+b)^{2}=a^{2}+2ab+b^{2} | (a-b)^{2}=a^{2}-2ab+b^{2} |
Second formula
(a^{2}-b^{2})=(a+b)(a-b)
Third formula
Positive sign | Negative |
a^{3}+b^{3}=(a+b)^{3}-3ab(a+b) | a^{3}-b^{3}=(a-b)^{3}+3ab(a-b) |
based on above formulas, we can derive some more formulas for example
(a+b)^{2}=a^{2}+2ab+b^{2}=> a^{2}+b^{2}=(a+b)^{2}-2ab
Same way,
a^{3}+b^{3}=(a+b)^{3}-3ab(a+b)
=> (a+b)^{3}= a^{3}+b^{3}+3ab(a+b)= a^{3}+b^{3}+3ab^{2}+3a^{2}b
Trigonometry: Formulas
Again, only three formulas
- Sin^{2}a+cos^{2}a=1
- Sec^{2}a-tan^{2}a=1
- Cosec^{2}a-cot^{2}a=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: Sin^{2}a+cos^{2}a=1 |
Cosec and sec | Here cosec comes with cot (and not Sec). Cosec^{2}a-cot^{2}a=1 |
Tan and cot | Here Tan comes with sec (and not Cot): Sec^{2}a-tan^{2}a=1 |
Based on these formula, you can derive many other formulas for example
Tan=sin/cos.
And we now know that Sin^{2}a+cos^{2}a=1. So
cos^{2}A=1-sin^{2}a
cosA=sq.root of (1-sin^{2}a)
now pluggin this back into Tan’s ratio
TanA=sinA/cosA
=sinA/ sq.root(1-sin^{2}A)
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 sin^{2}a+[1/(1+tan^{2}a)]
We know the Sec^{2}a-tan^{2}a=1=>sec^{2}a=1+tan^{2}a
Substituting this in the question
sin^{2}a+1/(1+tan^{2}a)
=sin^{2}a+(1/sec^{2}a) ; but cos= 1/sec
= Sin^{2}a+cos^{2}a
=1
Q. if 3cos^{2}A+7sin^{2}A=4 then find value of cotA, given that A is an acute angle?
Approach
Instead of 7sin^{2}a=3sin^{2}a+4sin^{2}a (because 4+3=7)….eq1
Now let’s look at the question
3cos^{2}a+7sin^{2}a=4
3cos^{2}a+3sin^{2}a+4sin^{2}a =4 ; using value from eq1
3(cos^{2}a+sin^{2}a) +4sin^{2}a =4
3(1)+ 4sin^{2}a =4
4sin^{2}a=4-3
Sin^{2}A=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 cos^{2}a+cos^{4}a=1, what is the value of tan^{2}a+tan^{4}a
Approach
We know that Sin^{2}a+cos^{2}a=1 =>sin^{2}a=1-cos^{2}a…eq1
In the question, we are given that
cos^{2}a+cos^{4}a=1
taking cos^{2}a on the right hand side
cos^{4}a=1-cos^{2}a
taking value from eq1
cos^{4}a=sin^{2}a
cos^{2}a x cos^{2}a = sin^{2}a ; laws of surds and indices
cos^{2}a=(sin^{2}a/cos^{2}a)
cos^{2}a=tan^{2}a……eq2
now lets move to the next part. in the question, we have to find the value of
tan^{2}a+tan^{4}a
=tan^{2}a+(tan^{2}a)^{2} ; because (a^{2})^{2}=a^{2×2=4}
= cos^{2}a+cos^{4}a ;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 sin^{4}a-cos^{4}a
We know that
(a^{2}-b^{2})=(a+b)(a-b)
So instead of sin^{4}a-cos^{4}a, I can write
(sin^{2}a)^{2}-(cos^{2}a)^{2} ; because a^{4=2×2} =(a^{2})^{2}
=(Sin^{2}a+cos^{2}a)(Sin^{2}a-cos^{2}a)
=1 x (Sin^{2}a-cos^{2}a) ; because (a^{2}-b^{2})=(a+b)(a-b)
=(Sin^{2}a-cos^{2}a)
We can further simplify this as (sina-cosa)x(sina-cosb)
Q. if sin^{4}a-cos^{4}a=-2/3 then what is the value of 2cos^{2}a-1
Approach
We know that Sin^{2}a+cos^{2}a=1=> sin^{2}a=1- cos^{2}a….(eq1)
From the previous sum, we already know that
LHS | RHS |
sin^{4}a-cos^{4}a= | (sin^{2}a-cos^{2}a) |
(1-cos^{2}a-cos^{2}a) ; applying eq.1 | |
-2/3 (given in the Qs.) | =1-2cos^{2}a |
2/3 | 2cos^{2}a-1 ; multiplying both sides with (-1) |
Therefore, 2cos^{2}a-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}=a^{2}-2ab+b^{2 }on each of them
(coseca-sina)^{2} | +(seca-cosa)^{2} | -(cota-tana)^{2} |
Cosec^{2}a-2cosecA x sinA+sin^{2}a | +sec^{2}a-2secA x cosA+cos^{2}a | -(cot^{2}A-2cotA x tanA +tan^{2}A) |
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:
Cosec^{2}a-2+sin^{2}a | +sec^{2}a-2+cos^{2}a | -(cot^{2}A-2 +tan^{2}A) |
=Cosec^{2}a-2+sin^{2}a+ sec^{2}a-2+cos^{2}a- cot^{2}A+2 -tan^{2}A ;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.sin^{2}a+cos^{2}a=1)
= (Cosec^{2}a- cot^{2}A) +(sin^{2}a+ cos^{2}a)+( sec^{2}a -tan^{2}A)+(-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}=a^{2}-2ab+b^{2}
LHS | RHS |
(sinA-cosA)^{2}= | Sin^{2}a-2sinA x cosA +cos^{2}A |
(7/13)^{2}= | 1-2sinA x cosA; because Sin^{2}A+cos^{2}A=1 |
Therefore, 2sinA x cosA=1-(7/13)^{2}=….eq1
Moving forward, we also know that
(a+b)^{2}=a^{2}+2ab+b^{2}
LHS | RHS |
(sinA+cosA)^{2}= | Sin^{2}a+2sinA x cosA +cos^{2}A |
1+2sinA x cosA; because Sin^{2}A+cos^{2}A=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 Cos^{2}a+(1/1+cot^{2}a)
- 0
- 1
- 2
- 3
- Find value of 2sin^{2}a+4sec^{2}a+5cot^{2}a+2cos^{2}a-4tan^{2}a-5cosec^{2}a
- 0
- 1
- 2
- 3
- Find the value of (cosA-sinA)^{2}+(cosA+sinA)^{2}
- 0
- 1
- 2
- 3
- Find the value of Cot^{2}A x (sec^{2}A-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
- Cos^{6}A+sin^{6}A=
- 1-3(cosA x sinA)^{2}
- 1+3(cosA x sinA)^{2}
- 1-3(cosA x sinA)^{3}
- 1-3(cosA x sinA)^{6}
- (sin^{2}A x cos^{2}B) – (cos^{2}A x sin^{2}B) will be equal to
- Sin^{2}A-cos^{2}A
- Sin^{2}A+cos^{2}A
- Cos^{2}A – cos^{2}A
- Sin^{2}A-Sin^{2}B
- (tanA+cotA)(secA-cosA)(cosecA-sinA)=
- 0
- 1
- 2
- 3
- If Tan^{2}A+Tan^{4}A=1 then what is the value of cos^{2}A+cos^{4}A, 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
- [(sin^{2}20+sin^{2}70)/(sec^{2}50-cos^{2}40)]+2cosec^{2}58-2(cot58xtan32)-(4tan13 x tan37 x tan45 x tan53 x tan77)
- 0
- -1
- 1
- None of above
- If cotA=root7, what is the value of (cosec^{2}A-sec^{2}A)/( cosec^{2}A+sec^{2}A)
- 3/2
- 3/4
- 4/3
- 2/3
- (sinA+cosecA)^{2}+(cosA+secA)^{2} will be equal to
- Tan^{2}A+cot^{2}A+4
- Tan^{2}A-cot^{2}A+5
- Tan^{2}A-cot^{2}A
- Tan^{2}A+cot^{2}A+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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149 Comments on “[Trigonometry] Type#4: Questions based on Trig and Algebra combo”
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
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