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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.

user3002473
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user2936034
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  • 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.

     – user2936034 Oct 08 '14 at 12:14

3 Answers3

4

Hints

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

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

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

C.S.
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3

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.
Hakim
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  • 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. :(

     – user2936034 Oct 08 '14 at 12:32
  • @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? – Hakim Oct 08 '14 at 12:35
  • 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?

     – user2936034 Oct 08 '14 at 12:45
  • @user2936034 Yes, now do the same things for (2) and (3). – Hakim Oct 08 '14 at 12:46
  • Thanks! I'll proceed to number 2 and 3. – user2936034 Oct 08 '14 at 12:50
  • @user2936034 You're welcome, glad I could help! :-) – Hakim Oct 08 '14 at 12:51
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    I solved all the problems now. :-) I really appreciate your help. – user2936034 Oct 08 '14 at 13:30
  • @user2936034 Keep up with the great work! ;-) – Hakim Oct 08 '14 at 13:31
  • Question: How $$ \frac{1}{cos^2x} + \frac{1}{sin^2x} $$

    becomes

    $$ \frac{cos^2x + sin^2x}{cos^2x sin^2x} $$

     – user2936034 Oct 08 '14 at 15:19
  • @user2936034 By unifying the denominator we get: $$\dfrac1{\cos^2x}+\dfrac1{\sin^2x}=\dfrac{\sin^2x}{\cos^2x\sin^2x}+\dfrac{{{{{{{{{{{{{{{\cos^{2}x}}}}}}}}}}}}}}}{\cos^2x\sin^2x}=\dfrac{\cos^2x+\sin^2x}{\cos‌​^2x\sin^2x}.$$ – Hakim Oct 08 '14 at 15:23
  • Oh, I see. You can switch $ sin^2x + cos^2x $ to $ cos^2x + sin^2x $? – user2936034 Oct 08 '14 at 15:30
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    @user2936034 Yes, $a+b=b+a$ and $a\cdot b=b\cdot a$ for all 'numbers' $a$ and $b$. – Hakim Oct 08 '14 at 15:32
0

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 $$\frac{c^2}{b^2}+\frac{c^2}{a^2}=\frac{c^2}{a^2}\cdot\frac{c^2}{b^2}$$ $$\csc^2\theta+\sec^2\theta=\sec^2\theta\csc^2\theta$$

John Joy
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