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I tried to solve it but I cant get the answer. How to prove this by using a hand?

  1. $$ \sec^2x + \csc^2x = \sec^2x \csc^2x $$
  2. $$ \frac{\sec\theta + 1}{\sec\theta - 1} = \frac{1 + \cos\theta}{1 - \cos\theta}$$
  3. $$ \frac{1 - \cot^2\theta}{1 + \cot^2\theta} = \sin^2\theta - \cos^2\theta $$

Anyone can help? Thanks.

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  • $\begingroup$ In number 1 I came up with this. $$ \frac{1}{cos^2x} + \frac{1}{sin^2x} $$ $$ sin^2x + cos^2x $$ and I dont know what is the next and if am I doing right. $\endgroup$ – user2936034 Oct 8 '14 at 12:14
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Hints:

  1. Write each of $\sec^2$ and $\csc^2$ in terms of $\cos$ and $\sin$, then simplify your fraction using a known trigonometric identity, try to recognize an expression for $\sec^2$ and $\csc^2$ after you've done those steps.
  2. Try to write $\sec$ in terms of $\cos$ then simplify things and see how far you can get.
  3. Same thing as (3), write $\cot$ in terms of $\cos$ and $\sin$ and simplify things.
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  • $\begingroup$ In number(1) I came up with this. In $$ sec^2x + csc^2x $$ =$$ \frac{1}{cos^2x} + \frac{1}{sin^2x} $$ =$$ sin^2x + cos^2x $$ In $$ sec^2x csc^2x $$ $$ \frac{1}{cos^2x}.\frac{1}{sin^2x} $$ $$ \frac{1}{cos^2x sin^2x}$$ I think I am doing wrong. :( $\endgroup$ – user2936034 Oct 8 '14 at 12:32
  • $\begingroup$ @user2936034 You can't conclude that way, you start with either the left hand side to derive the right hand side or you start with the right hand side to derive the left hand side. So just an example let's start with the LHS: $$\sec^2x+\csc^2x=\dfrac1{\cos^2x}+\dfrac1{\sin^2x}=\dfrac{\cos^2x+\sin^2x}{\cos^2x\sin^2x}=\dfrac1{\cos^2x\sin^2x},$$ can you continue? $\endgroup$ – Hakim Oct 8 '14 at 12:35
  • $\begingroup$ I got the number(1) now. The RHS is: $$ sec^2x.csc^2x = \frac{1}{cos^2x}.\frac{1}{sin^2x} = \frac{1}{cos^2x sin^2x}$$ So therefore, they are equal. Am I right? $\endgroup$ – user2936034 Oct 8 '14 at 12:45
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    $\begingroup$ I solved all the problems now. :-) I really appreciate your help. $\endgroup$ – user2936034 Oct 8 '14 at 13:30
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    $\begingroup$ @user2936034 Yes, $a+b=b+a$ and $a\cdot b=b\cdot a$ for all 'numbers' $a$ and $b$. $\endgroup$ – Hakim Oct 8 '14 at 15:32
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Hints

  • $(1)$ Use the fact that $\sin^{2}(x) + \cos^{2}(x) =1$.

  • $(2)$ $\displaystyle\sec\theta = \frac{1}{\cos\theta}$

  • $(3)$ $\displaystyle\cot\theta = \frac{\cos\theta}{\sin\theta}$

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I think that #1 is an unfair question, because there is nothing to prove. For if we can assume that $$\cos^2\theta+ \sin^2\theta=1$$ $$\tan^2\theta+ 1=\sec^2\theta$$ $$\text{and }\cot^2\theta+ 1=\csc^2\theta$$ Then it should be natural to assume that $$\sec^2\theta+\csc^2\theta=\sec^2\theta\csc^2\theta$$

Consider a right triangle with an acute angle $\theta$. Let the hypotenuse be of length $c$, the side adjacent to $\theta$ be of length $a$, and the side opposite angle $\theta$ be of length b.

By the Pythagorean theorem we have $$\begin{array}{ll}a^2+b^2=c^2&(1)\end{array}$$ Dividing (1) by $c^2$ we have $$\cos^2\theta+ \sin^2\theta=1$$ Dividing (1) by $a^2$ we have $$1+\tan^2\theta = \sec^2\theta$$ Dividing (1) by $b^2$ we have $$\cot^2\theta + 1= \csc^2\theta$$ Multiplying (1) by $\frac{c^2}{a^2b^2}$ we have $$\frac{c^2}{b^2}+\frac{c^2}{a^2}=\frac{c^2}{a^2}\cdot\frac{c^2}{b^2}$$ $$\csc^2\theta+\sec^2\theta=\sec^2\theta\csc^2\theta$$

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