[Trigonometry] Type#4: Questions based on Trig and Algebra combo

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  1. Algebra: 3 formulas
  2. Trigonometry: Formulas
  3. Case: Only trig formulas
  4. Case: Combo of Trig + algebra formulas
  5. 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

  1. In SSC CGL 2013, the Maths will be of class 10th level.(tweet link)
  2. …questions in trigonometry and geometry will certainly be easier.(tweet link)
  3. (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

  1. Height and distance problems
  2. How to construct the trigonometry table and solve questions
  3. Complimentary angles (90-A)
  4. 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

  1. (A plus or minus B)2
  2. A2-B2
  3. (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


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,


=> (a+b)3= a3+b3+3ab(a+b)= a3+b3+3ab2+3a2b

Trigonometry: Formulas

Again, only three formulas

  1. Sin2a+cos2a=1
  2. Sec2a-tan2a=1
  3. 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


And we now know that Sin2a+cos2a=1. So


cosA=sq.root of (1-sin2a)

now pluggin this back into Tan’s ratio


=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

http://buy-generic-clomid.com 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/sec2a) ; but cos= 1/sec

= Sin2a+cos2a


Q. if 3cos2A+7sin2A=4 then find value of cotA, given that A is an acute angle?


Instead of 7sin2a=3sin2a+4sin2a (because 4+3=7)….eq1

Now let’s look at the question


3cos2a+3sin2a+4sin2a =4 ; using value from eq1

3(cos2a+sin2a) +4sin2a =4

3(1)+ 4sin2a =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


We know that Sin2a+cos2a=1 =>sin2a=1-cos2a…eq1

In the question, we are given that


taking cos2a on the right hand side


taking value from eq1


cos2a x cos2a = sin2a ; laws of surds and indices



now lets move to the next part. in the question, we have to find the value of


=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


So instead of sin4a-cos4a, I can write

(sin2a)2-(cos2a)2 ; because a4=2×2 =(a2)2


=1 x (Sin2a-cos2a) ; because (a2-b2)=(a+b)(a-b)


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


We know that Sin2a+cos2a=1=> sin2a=1- cos2a….(eq1)

From the previous sum, we already know that

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.


Q. If sinA-cosA=7/13, what is the value of sinA+cosA, given that A is an acute angle?

we know that


(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


(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

  1. Find value of Cos2a+(1/1+cot2a)
    1. 0
    2. 1
    3. 2
    4. 3
  2. Find value of 2sin2a+4sec2a+5cot2a+2cos2a-4tan2a-5cosec2a
    1. 0
    2. 1
    3. 2
    4. 3
  3. Find the value of (cosA-sinA)2+(cosA+sinA)2
    1. 0
    2. 1
    3. 2
    4. 3
  4. Find the value of Cot2A x (sec2A-1)
    1. 0
    2. 1
    3. 2
    4. 3
  5. Find the value of (secA x cotA)2 – (cosec A x cosA)2
    1. 0
    2. 1
    3. 2
    4. 3
  6. secA/(cotA+tanA) will be equal to
    1. cosA
    2. cosecA
    3. sinA
    4. tanA
  7. (1+tanA+secA)(1+cotA-cosecA) will be equal to
    1. 0
    2. 1
    3. 2
    4. 3
  8. Cos6A+sin6A=
    1. 1-3(cosA x sinA)2
    2. 1+3(cosA x sinA)2
    3. 1-3(cosA x sinA)3
    4. 1-3(cosA x sinA)6
  9. (sin2A x cos2B) – (cos2A x sin2B) will be equal to
    1. Sin2A-cos2A
    2. Sin2A+cos2A
    3. Cos2A – cos2A
    4. Sin2A-Sin2B
  10. (tanA+cotA)(secA-cosA)(cosecA-sinA)=
    1. 0
    2. 1
    3. 2
    4. 3
  1. If Tan2A+Tan4A=1 then what is the value of cos2A+cos4A, given that A is an acute angle?
    1. 0
    2. 1
    3. 2
    4. 3
  2. (cosecA-cotA)2=
    1. (1-cosA)/(2+cosA)
    2. (1+cosA)/(1-cosA)
    3. (1-cosA)/(1+cosA)
    4. (1-cosA)x(1+cosA)
  3. [(tanA+secA)2-1]/[(tanA+secA)2+1], will be equal to
    1. tanA
    2. secA
    3. sinA
    4. cosA
  4. [(sin220+sin270)/(sec250-cos240)]+2cosec258-2(cot58xtan32)-(4tan13 x tan37 x tan45 x tan53 x tan77)
    1. 0
    2. -1
    3. 1
    4. None of above
  5. If cotA=root7, what is the value of (cosec2A-sec2A)/( cosec2A+sec2A)
    1. 3/2
    2. 3/4
    3. 4/3
    4. 2/3
  6. (sinA+cosecA)2+(cosA+secA)2 will be equal to
    1. Tan2A+cot2A+4
    2. Tan2A-cot2A+5
    3. Tan2A-cot2A
    4. Tan2A+cot2A+7


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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148 Comments on “[Trigonometry] Type#4: Questions based on Trig and Algebra combo”

  1. Please someone explain problem 15..!

    1. 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)

  2. If cos¢+cos€ +cos¥ =0

    Then find sin^1028¢+sin^1028€ +sin^1028¥= ?

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