Some important Short tricks on Trigonometric identities for SSC CGLE (Staff Selection Commission Combined Graduate Level Examination) are as follows:

Pythagorean Identities
• sin2 θ + cos2 θ = 1
• tan2 θ + 1 = sec2 θ
• cot2 θ + 1 = csc2 θ
Negative of a Function
• sin (–x) = –sin x
• cos (–x) = cos x
• tan (–x) = –tan x
• csc (–x) = –csc 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 = Csc B

For example:
If tan (x+y) tan (x-y) = 1, then find tan (2x/3)?
Solution:
Tan A = Cot B, 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, (A › B) Then
• Sin A = Cos B
• Cos A = – Sin B
• Tan A = – Cot B
If A ± B = 180o, then
• Sin A = Sin B
• Cos A = – Cos B
If A + B = 180o
Then, tan A = – tan B
If A – B = 180o
Then, tan A = tan B

For example:
Find the Value of tan 80o + tan 100o ?
Solution: Since 80 + 100 = 180
Therefore, tan 80o + tan 100o = 1
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 2 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 & csc θ – sec θ = q
then P – (1/p) = 2/q

For Example:
If sin θ + cos θ = 2 , then find the value of csc θ – sec θ:
Solution:
By using formulae:
P – (1/p) = 2/q
2-(1/2) = 3/2 = 2/q
Q = 4/3 or csc θ – sec θ = 4/3
If
a cot θ + b csc θ = m & a csc 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 csc2x + 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 csc2 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 csc 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

Attached Files

Trigonometry Question Paper for SSC CGL Exam.pdf (446.0 KB, 102 views)