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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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closed as too broad by 900 sit-ups a day, Ivo Terek, Tomás, anorton, RecklessReckoner Aug 1 at 3:15

There are either too many possible answers, or good answers would be too long for this format. Please add details to narrow the answer set or to isolate an issue that can be answered in a few paragraphs.If this question can be reworded to fit the rules in the help center, please edit the question.

Maybe you should tell us what you have tried? –  ᴊ ᴀ s ᴏ ɴ Sep 23 '12 at 6:10
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
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


$1$. 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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