Trig problems giving a hard time.

[labin.ojha]

I was having some free time and decided to do some mathematics from my high school mathematics book.These two problems remained, and I am completely clueless to the solution approach.

Homework Statement

A. If sin(θ)-cos(θ)=1, prove that sin(θ)+cos(θ)=±1
B. If tan(θ)+sec(θ)=10, prove that sin^{2}(θ)+cos^{2}(θ)=1

Homework Equations

The Attempt at a Solution

The approach to both problems were similar, I squared both sides of the given equations, and used trig identities at an attempt of simplifying.

That got me nowhere.

Pointers would be helpful.

 

[haruspex]

A. If sin(θ)-cos(θ)=1, prove that sin(θ)+cos(θ)=±1
B. If tan(θ)+sec(θ)=10, prove that sin^{2}(θ)+cos^{2}(θ)=1
I squared both sides of the given equations, and used trig identities at an attempt of simplifying.

That certainly works for A. If you still can't see it, please post your working.
For B, pay close attention to what is to be proved.

 

[labin.ojha]

That certainly works for A. If you still can't see it, please post your working.
For B, pay close attention to what is to be proved.

Found a way for A:

sin(θ)-cos(θ)=1
[Squaring]
sin^{2}(θ)-2sin(θ)cos(θ)+cos^{2}(θ)=1=sin^{2}(θ)+cos^{2}(θ)
sin^{2}(θ)+cos^{2}(θ)=sin^{2}(θ)+2sin(θ)cos(θ)+cos^{2}(θ)
1=(sin(θ)+cos(θ))^{2}
[Taking square roots]
sin(θ)+cos(θ)=\pm1

For B, the 'to prove' equation is an identity but getting it from the given expression is being a problem
because as I square the both sides , it gets messier and hopeless.

 

[Dick]

Found a way for A:

sin(θ)-cos(θ)=1
[Squaring]
sin^{2}(θ)-2sin(θ)cos(θ)+cos^{2}(θ)=1=sin^{2}(θ)+cos^{2}(θ)
sin^{2}(θ)+cos^{2}(θ)=sin^{2}(θ)+2sin(θ)cos(θ)+cos^{2}(θ)
1=(sin(θ)+cos(θ))^{2}
[Taking square roots]
sin(θ)+cos(θ)=\pm1

For B, the 'to prove' equation is an identity but getting it from the given expression is being a problem
because as I square the both sides , it gets messier and hopeless.

If the equation to be proved is an identity, you don't have to get it from the other expression. It's just plain always true.

 

[labin.ojha]

If the equation to be proved is an identity, you don't have to get it from the other expression. It's just plain always true.

Yes, it is. But I'll look for the solution and post it here as soon as i get it .

EDIT:
the question seems to have been removed from the new edition of the book, mine was an old one.

 

[Dick]

Yes, it is. But I'll look for the solution and post it here as soon as i get it .

There's really nothing to look for or get. This is exactly like proving "If x=2 then x=x." x=x is true regardless of whether x=2 is true. So "If x=2 then x=x." is a true statement.

 

[Dick]

EDIT:
the question seems to have been removed from the new edition of the book, mine was an old one.

Removing it is a good idea. In the context of proving trig stuff, it's only going to cause confusion.

 

[labin.ojha]

I don't know about Question B, but for Q. A
got a total of three methods of doing it.

http://www.reddit.com/r/cheatatmathhomework/comments/1iy7dg/trig_problems_giving_a_hard_time/