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

(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

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

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=a^2×2=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 a^4=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

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

11. 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
12. (cosecA-cotA)²=
    1. (1-cosA)/(2+cosA)
    2. (1+cosA)/(1-cosA)
    3. (1-cosA)/(1+cosA)
    4. (1-cosA)x(1+cosA)
13. [(tanA+secA)²-1]/[(tanA+secA)²+1], will be equal to
    1. tanA
    2. secA
    3. sinA
    4. cosA
14. [(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
15. If cotA=root7, what is the value of (cosec²A-sec²A)/( cosec²A+sec²A)
    1. 3/2
    2. 3/4
    3. 4/3
    4. 2/3
16. (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

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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