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  1. \begin{align}\frac {{\sec{\theta }}+{\tan{\theta }}}{{\sec{\theta }}-{\tan{\theta }}}={{\left[{\frac {1+{\sin{\theta}}}{\cos{\theta }}}\right]}}^{{2}}\end{align} Prove this

  2. $ABC$ is a right triangle with right angle at $B$. $BC=7$ cm and $AC-AB=1$ cm. Find $\cos A - \sin A$.

  3. If $x = a \sec \theta + b \tan\theta$ and $y = a \tan\theta + b \sec \theta$, then prove that $x^2 - y^2 = a^2 - b^2$
  4. If $\frac x a \cos\theta + \frac y b \sin \theta = 1$ and $\frac x a \sin\theta - \frac y b \cos\theta = 1$, then prove that $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 2$.
  5. 2($sin^6\theta + cos^6\theta$) - 3($sin^4\theta + cos^4\theta$) + 1 = 0
  6. If tan A $\sqrt{2} - 1$. Show that SinA CosA = $\sqrt{2} \over 4$
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put on hold as too broad by 900 sit-ups a day, Ivo Terek, Tomás, anorton, RecklessReckoner yesterday

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Maybe you should tell us what you have tried? –  ᴊ ᴀ s ᴏ ɴ Sep 23 '12 at 6:10
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This isn't a question; it's an assignment! –  Blue Sep 23 '12 at 6:11
    
Pretty much a lot of different things...Tried to evaluate using different properties and taking different forms of the identities, but had absolutely no luck at all. These 6 are the ones i was able to solve... –  Aayush Agrawal Sep 23 '12 at 6:12
    
Also its not an assignment... Its from a practice book. –  Aayush Agrawal Sep 23 '12 at 6:13
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Sorry, @aayush. I've been misunderstood. My intention was to say that, given the number of items in your question, you were giving us an assignment. (And I should've included a smiley.) It's perfectly okay to seek advice on homework or practice questions (properly tagged). –  Blue Sep 23 '12 at 6:27

2 Answers 2

up vote 0 down vote accepted

Hints:

$1$. Express the left-hand side in terms of sines and cosines, and simplify. You should get after a while something like $\frac{1+\sin\theta}{1-\sin\theta}$. Now multiply top and bottom by something useful.

$2$. Let $x=AB$. Then $AC=x+1$. Now use the Pythagorean Theorem.

$3$. Calculate $(a\sec\theta+b\tan\theta)^2-(a\tan\theta+b\sec\theta)^2$ by expanding the squares. Then use the fact that $\sec^2\theta-\tan^2\theta=1$.

$4$. Square both expressions, expand the squares, and add. Something nice will happen.

$5$. This one may be kind of hard. Maybe $\sin^6\theta=(1-\cos^2\theta)\sin^4\theta$ and $\cos^6\theta=(1-\sin^2\theta)\cos^4\theta$. Add. We get $\sin^6\theta+\cos^6\theta=\sin^4\theta+\cos^4\theta-\sin^2\theta\cos^2\theta(\sin^2\theta+\cos^2\theta)$.

$6$. Make a right triangle with leg opposite $A$ equal to $\sqrt{2}-1$, and leg adjacent to $A$ equal to $1$. Then $\tan A$ is $\frac{\sqrt{2}-1}{1}$. Find the square of the hypotenuse. You now should be able to read off $\sin A\cos A$.

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I already reached $1+sin\theta \over 1-sin\theta$ but i dont get how i can possibly convert that to the RHS...I tried squaring but then then the denominator becomes 1 - 2sin$\theta$ + $sin\theta^2$. I tried sin theta but that gave $1 + sin\theta^2$ over cos theta. –  Aayush Agrawal Sep 23 '12 at 6:31
    
Second one also i still failed to do...I just cant do this one at all –  Aayush Agrawal Sep 23 '12 at 6:35
    
Multiply top and bottom by $1+\sin\theta$. The bottom becomes $1-\sin^2\theta$, which is $\cos^2\theta$. The end. –  André Nicolas Sep 23 '12 at 6:35
    
OMG, but how do you solve these just like that? I mean how do the answers just come to you? Is there some kind of general procedure? Because i know all the required facts but still cant come up with a solution, while others can just see it and immediately tell me one –  Aayush Agrawal Sep 23 '12 at 6:37
    
We get $x^2+49=(x+1)^2=x^2+2x+1$. So $x=24$. Now you can find the sine, cosine from a picture of the triangle (like $\cos$ is opposite divided by hypotenuse). –  André Nicolas Sep 23 '12 at 6:37

5). Use $a^3+b^3=(a+b)^3-3ab(a+b)$ and $a^2+b^2=(a+b)^2-2ab$ and note that $a+b=1$

6). $\sin A\cos A=\frac{\sin A}{\cos A}.\cos^2 A=\tan A \frac{1}{1+\tan^2 A}=\frac{\sqrt{2}-1}{1+(\sqrt{2}-1)^2}=\frac{\sqrt{2}-1}{2\sqrt{2}(\sqrt{2}-1)}=\frac{\sqrt{2}}{4}$

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