  # Important Notes & Short Tricks on Trigonometric Identities

By BYJU'S Exam Prep

Updated on: September 25th, 2023 Today we will be covering a very important topic from the Advance Maths part of the Quantitative Aptitude section that is – Important Notes & Short Tricks on Trigonometric Identities. These identities will help in most of the government competitive exams like SSC CGL, CHSL, IB ACIO, RRB NTPC etc.

Table of content ## Important Short Tricks on Trigonometric Identities

Pythagorean Identities

• sin2 θ + cos2 θ = 1
• tan2 θ + 1 = sec2 θ
• cot2 θ + 1 = cosec2 θ

Negative of a Function

• sin (–x) = –sin x
• cos (–x) = cos x
• tan (–x) = –tan x
• cosec (–x) = –cosec x
• sec (–x) = sec x
• cot (–x) = –cot x

If A + B = 90o, Then

• Sin A = Cos B
• Sin2A + Sin2B = Cos2A + Cos2B = 1
• Tan A = Cot B
• Sec A = Cosec B

For example:

If tan (x+y) tan (x-y) = 1, then find tan (2x/3)?

Solution:

If A+B = 90°, then

Tan A = Cot B or Tan A×Tan B = 1

So, A +B = 90o

(x+y)+(x-y) = 90o, 2x = 90o , x = 45o

Tan (2x/3) = tan 30o = 1/√3

If A – B = 90o, and (A › B) Then,

• Sin A = Cos B
• Cos A = – Sin B
• Tan A = – Cot B
• Cosec A = – Sec B
• Sec A = Cosec B

If A ± B = 180o, then

• Cos A = – Cos B

If A + B = 180o

Then,

• Sin A = Sin B
• tan A = – tan B

If A – B = 180o

Then,

• Sin A = – Sin B
• tan A = tan B

If A + B + C = 180o, then

Tan A + Tan B +Tan C = Tan A * Tan B *Tan C

sin θ * sin 2θ * sin 4θ = ¼ sin 3θ

cos θ * cos 2θ * cos 4θ = ¼ cos 3θ

For Example: What is the value of cos 20o cos 40o cos 60o cos 80o?

Solution: We know, cos θ * cos 2θ * cos 4θ = ¼ cos 3θ

Now, (cos 20o cos 40o cos 80o ) cos 60o

¼ (Cos 3*20) * cos 60o

¼ Cos2 60o = ¼ * (½)2 = 1/16

If  a sin θ + b cos θ = m     &    a cos θ – b sin θ = n

then a2 + b2 = m2 + n2

For Example:

If 4 sin θ + 3 cos θ = 2 , then find the value of  4 cos θ – 3 sin θ:

Solution:

Let 4 cos θ – 3 sin θ = x

By using formulae a2 + b2 = m2 + n2

42 + 32 = 22 + x2

16 + 9 = 4 + x2

X = √21

If

sin θ +  cos θ = p     &     cosec θ + sec θ = q

then p – (1/p) = 2/q

For Example:

If sin θ + cos θ = √2 , then find the value of  cosec θ + sec θ:

Solution:

By using formulae:

P – (1/p) = 2/q

√2-(1/√2) = 1/√2 = 2/q

q = 2√2 or cosec θ + sec θ = 2√2

If

a cot θ + b cosec θ = m     &    a cosec θ + b cot θ = n

then b2 – a2  = m2 – n2

If

cot θ + cos θ = x     &    cot θ – cos θ = y

then x2 – y2 = 4 √xy

If

tan θ + sin θ = x     &    tan θ – sin θ = y

then x2 – y2 = 4 √xy

If

y = a2 sin2x + b2 cosec2x + c

y = a2 cos2x + b2 sec2x + c

y = a2 tan2x + b2 cot2x + c

then,

ymin = 2ab + c

ymax = not defined

For Example:

If y = 9 sin2 x + 16 cosec2 x + 4 then ymin is:

Solution:

For, y min = 2* √9 * √16 + 4

= 2*3*4 + 20 = 24 + 4 = 28

If

y = a sin x + b cos x + c

y = a tan x + b cot x + c

y = a sec x + b cosec x + c

then,     ymin = – [√(a2+b2)] + c

ymax = + [√(a2+b2)] + c

For Example:

If y = 1/(12sin x + 5 cos x +20) then ymax is:

Solution:

For, y max = 1/x min

= 1/[- (√122 +52) +20] = 1/(-13+20) = 1/7

Sin2 θ, maxima value = 1, minima value = 0

Cos2 θ, maxima value = 1, minima value = 0

Here are some important questions of Trigonometric identities.

(1)Value of is

(a) (b) (c) (d)None of these

Ans.(a) is equal to (2)If is acute and then is equal to

(a)

(b)3

(c)  2

(d)  4

Ans.  (c)

If sum of the inversely proportional value is 2 so =2

or we can put = 45°

(3)The simplified value of(Secx Secy + tanx tany)2 – (Secx tany + tanx Secy)2 is

(a)-1

(b)0

(c)sec2x

(d)1

Ans. (d)

The simplified value of (Secx Secy + tanx tany)2 – (Secx tany + tanx Secy)is obtained by putting x = y = 45°

= (√2×√2 + 1×1)2 – (√2×1 + 1×√2)2

= (2+1)2 – (√2+√2)2

=(3)2 – (2√2)2

= 9 – 8 = 1

(4) Find the value of (a)  1

(b)  -1

(c)  2

(d)  -2

Ans. (c)

put  (5) If then is equal

(a)7/4

(b) 7/2

(c)5/2

(d)5/4

Ans. (d) as we know that  on solving we get sec = 5/4

Note: if x+y=a

and x-y=b

then x=(a+b)/2 and y=(x-y)/2

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