# Thread: Help on Trigonometric sums. (Assorted type)

1. ## Help on Trigonometric sums. (Assorted type)

I have some questions and doubts in Trigonometry. I hope somebody can solve these questions.

Q1) If cosec theeta - cot theeta = 1/2, o< theeta< pie/2, then cos theeta is equal to

a) 5/3 b) 3/5 c) -3/5 d) -5/3

I will post the following questions soon.

P.S. I would also like to have the steps to solve the problem please.

2. ## Re: Help on Trigonometric sums. (Assorted type)

Q2) If cosec theeta - sin theeta = a^3 and sec theeta - cos theeta = b^3, then (a^2 b^2)(a^2 + b^2) =

a) 1 b) -1 c) 2 d) none of these

3. ## Re: Help on Trigonometric sums. (Assorted type)

Originally Posted by surajkrishna
Q2) If xosec theeta - sin theeta = a^3 and sec theeta - cos theeta = b^3, then (a^2 b^2)(a^2 + b^2) =

a) 1 b) -1 c) 2 d) none of these
what is xo ?

4. ## Re: Help on Trigonometric sums. (Assorted type)

Originally Posted by surajkrishna
Q1) If xosec theeta - cot theeta = 1/2, o< theeta< pie/2, then cos theeta is equal to

a) 5/3 b) 3/5 c) -3/5 d) -5/3
I assume xosec means cosec. If I could have a pie every time I wrote a pi, I would be rich!

To solve this equation exactly, express cosec and cot through sin and cos, add the two fractions in the left-hand side and express $\sin\theta$ as $\sqrt{1-\cos^2\theta}$. In general, $\sin\theta=\pm\sqrt{1-\cos^2\theta}$, but since $0<\theta<\pi/2$, both $\sin\theta$ and $\cos\theta$ are positive. You'll get a quadratic equation in $\cos\theta$.

With these answer options, it is easy to eliminate three of them. As I said, $\cos\theta$ must be positive for $\theta$ in this range, and $\cos\theta$ cannot exceed 1.

5. ## Re: Help on Trigonometric sums. (Assorted type)

Originally Posted by surajkrishna
Q2) ... (a^2 b^2)
Syntax error: operator missing.

6. ## Re: Help on Trigonometric sums. (Assorted type)

Hello, surajkrishna!

$\text{Q1) If }\csc\theta - \cot\theta \,=\,\tfrac{1}{2},\,0<\theta< \tfrac{\pi}{2},\text{ find }\cos\theta$

. . $(a)\;\tfrac{5}{3}\quad(b)\;\tfrac{3}{5} \quad(c)\;\text{-}\tfrac{3}{5} \quad (d)\;\text{-}\tfrac{5}{3}$

We have: . $\frac{1}{\sin\theta} - \frac{\cos\theta}{\sin\theta} \:=\:\frac{1}{2} \quad\Rightarrow\quad \frac{1-\cos\theta}{\sin\theta} \:=\:\frac{1}{2} \quad\Rightarrow\quad 2(1-\cos\theta) \:=\:\sin\theta$

Square both sides: . $\big[2(1-\cos\theta)\big]^2 \:=\:\sin^2\!\theta \quad\Rightarrow\quad 4 - 8\cos\theta + 4\cos^2\!\theta \:=\:1-\cos^2\!\theta$

. . $5\cos^2\!\theta - 8\cos\theta + 3 \:=\:0 \quad\Rightarrow\quad (\cos\theta - 1)(5\cos\theta - 3) \:=\:0$

We have:

. . $\cos\theta -1 \:=\:0 \quad\Rightarrow\quad \cos\theta \:=\:1 \quad\Rightarrow\quad \theta \:=\:0\;\text{ . . . not in the domain}$

. . $5\cos\theta - 3 \:=\:0 \qiad\Rightarrow\quad \cos\theta \:=\:\tfrac{3}{5}\;\text{ . . . answer (b)}$

7. ## Re: Help on Trigonometric sums. (Assorted type)

Originally Posted by surajkrishna
I have some questions and doubts in Trigonometry. I hope somebody can solve these questions.

Q1) If cosec theeta - cot theeta = 1/2, o< theeta< pie/2, then cos theeta is equal to

a) 5/3 b) 3/5 c) -3/5 d) -5/3
It isn't really necessary to solve any equation here.

a)5/3, c) -3/5, and d) -5/3 are obviously impossible (do you see why?) so b) 3/5 is the only possible answer.
If you want to check, cosec= 5/4 and cot= 3/4. Their difference is 5/4- 3/4= 2/4= 1/2.

I will post the following questions soon.

P.S. I would also like to have the steps to solve the problem please.
Did you even try this? It's pretty close to being trivial.

8. ## Re: Help on Trigonometric sums. (Assorted type)

Q2) If cosec $\theta$ - sin $\theta$ = a3 and sec $\theta$ - cos $\theta$ = b3, find a2 x b2 (a2 + b2)

a) 1 b) -1 c) 2 d) none of these

9. ## Re: Help on Trigonometric sums. (Assorted type)

Originally Posted by surajkrishna
Q2) If cosec $\theta$ - sin $\theta$ = a3 and sec $\theta$ - cos $\theta$ = b3, find a2 x b2 (a2 + b2)

a) 1 b) -1 c) 2 d) none of these
Yes, sorry, I should have guessed that $a^2 b^2$ means multiplication.

Substituting $\theta=\pi/4$ gives $a^2b^2(a^2+b^2)=1$, so the answer is (a). It's possible to prove that the result is 1 for every $\theta$ by expressing cosec and sec through sin and cos and then honestly computing $a^2b^2(a^2+b^2)$. It is not too hard.

10. ## Re: Help on Trigonometric sums. (Assorted type)

Originally Posted by emakarov
Yes, sorry, I should have guessed that $a^2 b^2$ means multiplication.

Substituting $\theta=\pi/4$ gives $a^2b^2(a^2+b^2)=1$, so the answer is (a). It's possible to prove that the result is 1 for every $\theta$ by expressing cosec and sec through sin and cos and then honestly computing $a^2b^2(a^2+b^2)$. It is not too hard.
Can you please show the steps by expressing cosec and sec through sin and cos like the way you have told. I have tried it but I didn't get it.

11. ## Re: Help on Trigonometric sums. (Assorted type)

Originally Posted by surajkrishna
Can you please show the steps by expressing cosec and sec through sin and cos like the way you have told.
Expressing cosec and sec through sin and cos falls below the difficulty level appropriate for this forum. That is to say, I am sure you have done this. If you would like help, please show your work to the point where you are stuck.