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

- 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