- 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-2sinA x cosA +^{2}acos^{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+2sinA x cosA +^{2}acos^{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}

- 1-3(cosA x sinA)
- (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

- Sin
- (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

- Tan

# 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

for more articles on trigonometry and aptitude, visit Mrunal.org/aptitude

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

someone plzz help with question 5

Plz help me for problm no.15

put angle A=45 degree

(SecA*cotA)2=( 1/cosA)*(cosA/cosA). Here cancle cosA then this wii become (1/cosA)2 this wiil became cosecA2 inthe same way finally u wiil get the formule cose(squre)A-cot(squre)A=1

(sec A x cot A)2 – (cosec A x cos A)2

[(1/cos^2 A) x (Cos^2 A/Sin^2 A)] – [(1/Sin^2 A) x Cos^2 A)

[1/Sin^2 A] – Cot^2 A

Cosec^2 A -Cot^2 A = 1

Answer : B

Solving the questions for the 1st time, hence the late reply!!

Please give me solution of given example

If sinπx/2=x^2-2x+2, then the value of x is

brother put a=45

Question 14 is incorrect

Its not incorrect . I will give you the solution . if you want

Its just a simple trick

convert all then angles more than 45

for example write sin2 (70) as sin2 (90-20)

and apply co function than it becomes

cos2 (20)

now sin2(20)+cos2(20) =1

(SecA*cotA)^2 – (cosecA*cosA)^2

(1/cosA*cosA/sinA)-(1/sinA.cosA)

(1/sinA)^2 – (cosA/sinA)^2

Cosec^2 A – cot^2 A

1

These all questions are very easy to do if you read the formulas written above carefully . I found all these questions very easy after doing that you may also try . Its just basic algebra and trigonometric .

(seca+cosa)(seca-cosa)-2 + cos^2a = tan^2a.sec^2a

plz explain ques 8

these all questions are so easy.

but u can’t solve without formulas because they are totally depends upon formulas,

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