I am preparing for the SSC exams and finding difficulty in the preparation of the Trigonometry subject. Can you please provide me some study material of Trigonometry subject and sample papers of Trigonometry Questions along with their Answers for the SSC exam preparation?

As you asking here I am providing Trigonometry Questions And Answers For SSC Exams preparation :

Question 1: Prove the following identity:


Solution:
LHS = (sec$\theta$ - cose$\theta$)(1 + tan$\theta$ + cot$\theta$)
= ($\frac{sin\theta - cos\theta}{sin\theta cos\theta}$)($\frac{sin\theta cos\theta + 1}{sin\theta cos\theta}$)
[Using identities, sec$\theta$ = $\frac{1}{cos\theta}$, cose$\theta$ = $\frac{1}{sin\theta}$, tan$\theta$ = $\frac{sin\theta}{cos\theta}$ and cot$\theta$ = $\frac{cos\theta}{sin\theta}$]
= $\frac{(sin\theta - cos\theta)(sin\theta\ cos\theta + 1)}{sin^2\theta\ cos^2\theta}$
= $\frac{(sin\theta - cos\theta)(sin\theta\ cos\theta + sin^2\theta + cos^2\theta)}{sin^2\theta\ cos^2\theta}$
[Using, $sin^2\theta + cos^2\theta$ = 1]





= tan$\theta$ sec $\theta$ - cot$\theta$cosc$\theta$
= RHS

Question 2:
Prove the following identity:


Solution:
LHS = 1 + 2sec2A tan2A - sec4A - tan4A

= 1 - (sec4A - 2sec2A tan2A + tan4A)





[Using identity, sec2A - tan2A = 1]

= 1 - 1

= 0

= RHS
Question 3: Prove the following identity:


Solution:
LHS = [$\frac{1 - sinA cosA}{cosA(secA - cosecA)}$][$\frac{ sin^2 A - cos^2A}{sin^3A + cos^3A}$]

= [$\frac{1 - sinA cosA}{cosA(\frac{1}{cosA} - \frac{1}{sinA})}$][$\frac{(sin A + cos A)(sin A - cos A)}{ (sin A + cos A)(sin^2 A - sin A cos A + cos^2A)}$]

[Using identities, sec$\theta$ = $\frac{1}{cos\theta}$, cose$\theta$ = $\frac{1}{sin\theta}$, sin2A - cos2A = (sin A + cos A)(sin A - cos A) and sin3A + cos3A = (sin A + cos A)(sin2 A - sin A cos A + cos2A)]

= [$\frac{1 - sinA cosA}{\frac{sin A - cosA}{sinA}}$][$\frac{sin A - cos A}{sin^2A - sin A cos A + cos^2A}$] [By cancelling common terms]

= sin A[$\frac{1 - sinA cosA}{sin A - cos A}$] * [$\frac{sin A - cos A}{1 - sinA cosA}$]

[Using identity sin2 A + cos2 A = 1]

= sin A (By cancelling common terms)

= RHS

Lets us solve some more trigonometric examples using their identities:
Solved Examples
Question 1: Prove the following identity:


Solution:
LHS = (sec$\theta$ - 1)2 - (tan$\theta$ - sin$\theta$)2

= ($\frac{1}{cos\theta}$ - 1)2 - ($\frac{sin\theta}{cos\theta}$ - sin$\theta$)2

= ($\frac{1 - cos\theta}{cos\theta}$)2 - $\frac{sin^2\theta}{cos^2\theta}$(1 - sin$\theta$ * $\frac{cos\theta}{sin\theta}$)2

= ($\frac{1 - cos\theta}{cos\theta}$)2 - (1 - cos$\theta$)2 $\frac{sin^2\theta}{cos^2\theta}$

= (1 - cos$\theta$)2 [$\frac{1}{cos^2\theta} - \frac{sin^2\theta}{cos^2\theta}$]




Question 2:
Prove the following identity:


Solution:
LHS = $\frac{tan^3 \theta }{1 + tan^2 \theta}$ + $\frac{cot^3 \theta }{1 + cot^2 \theta}$

= $\frac{tan^3\theta }{sec^2\theta}$ + $\frac{cot^3\theta }{cose^2\theta}$











Solved Examples
Question 1:


Solution:
If tan A + sin A = m .................. (1)
tan A - sin A = n ................... (2)
Step 1:
Adding (1) and (2)

Step 2:
Subtracting (2) from (1)

Now


[Using, (a + b)2 = a2 + b2 + 2ab and (a - b)2 = a2 + b2 - 2ab]



Question 2:
Solution:








= $\frac{1}{\sqrt 2 + 1}$ * $\frac{\sqrt 2 - 1}{\sqrt 2 - 1}$ cos $\theta$

= ($\sqrt 2$ - 1)cos $\theta$






Question 3:
If x sin3$\theta$ + y cos3$\theta$ = sin $\theta$ cos $\theta$ and x sin $\theta$ - y cos $\theta$ = 0. Prove that x2 + y2 = 1.

Solution:
x sin3$\theta$ + y cos3$\theta$ = sin $\theta$ cos $\theta$ ..................... (i)

x sin $\theta$ - y cos $\theta$ = 0 ...................... (ii)



from (i) and (iii)







y = sin$\theta$ .................. (iv)

From (iii) and (iv), x = cos$\theta$

=> x2 + y2 = sin2$\theta$ + cos2$\theta$


Question 4:
Solution:


[using identity 1 - cos 2A = 2sin2 A similarly, 1 - cos 8A = 2sin24A and 1 - cos 4A = 2sin22A ]

[Using identity, sin2A = 2sin A cos A]


Question 5:
Solution:
Step 1:


[Using identity, cos 3A = 4cos3A - 3cosA]

[Using, cos2A = $\frac{1 + cos2A}{2}$]


or

Step 2:
let A = 15

Attached Files

Trigonometry Questions And Answers For SSC Exams.doc (115.0 KB, 157 views)