This discussion is now closed.

Check out other Related discussions

Core Mathematics; Trigonometry; Exercise 6D

Need help with the following questions. Please explain all steps little by little.
1. Solve the following equation in the given interval:
sec^2[theta] = 3tan[theta], 0<(and equal to) [theta] <(and equal to) 360º

2. Given that x = sec[theta] + tan [theta], show that 1/x = sec[theta]-tan[theta]

3. Given that 2sec^2[theta] - tan^2[theta] = p show that cosec^2[theta] = p-1/p-2, P cannot equal 2.

Please explain each step. Thanks a lot.

Reply 1

Jordan7

8

Mohit_C

Need help with the following questions. Please explain all steps little by little.
1. Solve the following equation in the given interval:
sec^2[theta] = 3tan[theta], 0<(and equal to) [theta] <(and equal to) 360º

Please explain each step. Thanks a lot.

sec^2[x] = 3tan[x],
sec^2x=tan^2x +1
tan^2x + 1 = 3tanx
0=tan^2x - 3tanx + 1 y=tanx
0= y^2 - 3y + 1
erm quadratic formula...

Reply 2

steve2005

17

The factorisation contains an error.

2y x y = 2y^2

lol yh i jus noticed

Thats right.

Mohit_C

Need help with the following questions. Please explain all steps little by little.
1. Solve the following equation in the given interval:
sec^2[theta] = 3tan[theta], 0<(and equal to) [theta] <(and equal to) 360º

.

I worked as far as one angle. There are several other solutions

Theres an error in yours aswel:
You have got :

tan^2[x] + 1 = 3tan[x] thn you got
tan^2[x] - 3tan[x] - 1=0 shouldnt it be +1 ???

Reply 7

steve2005

17

Jordan7

Theres an error in yours aswel:
You have got :

tan^2[x] + 1 = 3tan[x] thn you got
tan^2[x] - 3tan[x] - 1=0 shouldnt it be +1 ???

I can't see the error. Which line is it in.

Reply 8

Jordan7

8

the 6th line of your working

Reply 9

steve2005

17

Jordan7

the 6th line of your working

Are you winding me up?

I have written plus one.

Reply 10

Jordan7

8

o ok sorry its just the vertical line of the + is very very faint, on myn anyway.

Reply 11

Mohit_C

OP

10

Thanks guys, need help with the remaining ones, especially the last one which doesn't seem to make any sense whatsoever to me.

Reply 12

Jordan7

8

Mohit_C

Need help with the following questions. Please explain all steps little by little.

13. Given that 2sec^2[theta] - tan^2[theta] = p show that cosec^2[theta] = p-1/p-2, P cannot equal 2.

Please explain each step. Thanks a lot.

2sec^2[theta] - tan^2[theta] = p
2(tan^2x + 1) - tan^2x = p
tan^2x + 2 = p

cosec^2x= 1+ cot^2x
cosec^2x = 1 + (1/tan^2x)
1 = tan^2x / tan^2x
cosec^2x= tan^2x/tan^2x + 1/tan^2x
cosec^2x= (tan^2x +1) / ( tan^2x)
cosec^2x= (p-1) / (p-2)

Reply 13

Mohit_C

OP

10

Thanks Jordan7. I have a little question. On question 2, where you had to find 1/x, the book used the idea of rationalizing....can this be done with trigonometric functions?

Core Mathematics; Trigonometry; Exercise 6D

Related discussions

Articles for you