Free Introduction to Trigonometry 01 Practice Test - 10th Grade 

$\text{We know that}tan{30}^{\circ }=\frac{1}{\sqrt{3}}.$

$\Rightarrow \frac{1-ta{n}^2{30}^{\circ }}{1+ta{n}^2{30}^{\circ }}=\frac{1-(\frac{1}{\sqrt{3}}{)}^2}{1+(\frac{1}{\sqrt{3}}{)}^2}$

$=\frac{1-\left(\frac{1}{3}\right)}{1+\left(\frac{1}{3}\right)}$

$=\frac{3-1}{3+1}$

$=\frac{2}{4}$ = $\frac{1}{2}=0.5$ 

The given expression is 

$(\frac{1+\tan \theta }{1+\cot \theta }{)}^2=(\frac{1+\tan \theta }{1+\frac{1}{\tan \theta }}{)}^2$
                 
$=(\frac{1+\tan \theta }{\frac{\tan \theta +1}{\tan \theta }}{)}^2$
                 
$=(\frac{\left(\tan \theta \right)(1+\tan \theta )}{\tan \theta +1}{)}^2$
               
$={\tan }^2\theta$

$\frac{3\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }$ = $\frac{5}{\sin \theta }$

$\frac{3{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }=\frac{5}{\sin \theta }$
$3{\sin }^2\theta +{\cos }^2\theta =5\cos \theta$

3(1 - ${\cos }^2\theta$) + ${\cos }^2\theta$ = 5 $\cos \theta$

2${\cos }^2\theta$ + 5$\cos \theta$ - 3 = 0 

2$\cos \theta$  [$\cos \theta$ + 3] - 1($\cos \theta$ + 3) = 0

($\cos \theta$ + 3) (2$cos\theta$ - 1) = 0

$\cos \theta$ = -3     or     $\cos \theta$ = $\frac{1}{2}$

Note that $\cos \theta$ = -3  is not possible as $-1\le \cos \theta \le 1$                     
 Thus, $\theta$ = ${60}^{\circ }$

$ta{n}^{2}A+co{t}^{2}A+2\text{can be written as}(1+ta{n}^{2}A)+(1+co{t}^{2}A)$

We know that, $1+ta{n}^{2}A=se{c}^{2}A\text{and}1+co{t}^{2}A=cose{c}^{2}A$

$\Rightarrow ta{n}^{2}A+co{t}^{2}A+2=se{c}^{2}A+cose{c}^{2}A$

$=\frac{1}{co{s}^{2}A}+\frac{1}{si{n}^{2}A}$

$=\frac{si{n}^{2}A+co{s}^{2}A}{co{s}^{2}Asi{n}^{2}A}...(si{n}^{2}A+co{s}^{2}A=1)$

$=\frac{1}{co{s}^{2}Asi{n}^{2}A}$

$=se{c}^{2}Acose{c}^{2}A$

We know that, ${\sin }^{2}A+{\cos }^{2}A=1.$
 $1){\cos }^{2}A-{\sin }^{2}A=(1-{\sin }^{2}A)-{\sin }^{2}A$
$=1-2{\sin }^{2}A$

$2){\cos }^{2}A-{\sin }^{2}A={\cos }^{2}A-(1-{\cos }^{2}A)$
$={\cos }^{2}A-1+{\cos }^{2}A$
$=2{\cos }^{2}A-1$

$A+B+C={180}^{\circ }$
(sum of angles of a triangle)

$⟹B+C={180}^{\circ }-A$

$\therefore \frac{B+C}{2}=\frac{{180}^{\circ }-A}{2}$

$\frac{B+C}{2}={90}^{\circ }-\frac{A}{2}$

Taking sine on both sides we get,

$\sin \left(\frac{B+C}{2}\right)=\sin ({90}^{\circ }-\frac{A}{2})$

$\therefore \sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}$

$\frac{\sin A-2{\sin }^{3}A}{2{\cos }^{3}A-cosA}=\frac{\sin A(1-2{\sin }^{2}A)}{\cos A(2{\cos }^{2}A-1)}$

$=\frac{\sin A({\sin }^{2}A+{\cos }^{2}A-2{\sin }^{2}A)}{\cos A(2{\cos }^{2}A-({\sin }^{2}A+{\cos }^{2}A\left)\right)}$

$=\frac{\sin A({\cos }^{2}A-{\sin }^{2}A)}{\cos A({\cos }^{2}A-{\sin }^{2}A)}$

$=\tan A$

$\frac{sin\theta +1-cos\theta }{cos\theta -1+sin\theta }$

Dividing the numerator and denominator by $cos\theta$ we get,

$\frac{\frac{sin\theta }{cos\theta }+\frac{1}{cos\theta }-\frac{cos\theta }{cos\theta }}{\frac{cos\theta }{cos\theta }-\frac{1}{cos\theta }+\frac{sin\theta }{cos\theta }}$

$=\frac{tan\theta +sec\theta -1}{1-sec\theta +tan\theta }$

$\Rightarrow \frac{(tan\theta +sec\theta )-(se{c}^2\theta -ta{n}^2\theta )}{1-sec\theta +tan\theta }....(1+{\tan }^2\theta ={\sec }^2\theta )$

$=\frac{(\tan \theta +\sec \theta )-(\sec \theta -\tan \theta )(\sec \theta +\tan \theta )}{1-\sec \theta +\tan \theta }$

$=\frac{(\tan \theta +\sec \theta )(1-\sec \theta +\tan \theta )}{(1-\sec \theta +\tan \theta )}$

$=\tan \theta +\sec \theta$

$=\frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta }$

$=\frac{\sin \theta +1}{\cos \theta }$

[Multiplying the numerator and denominator by $(1-\sin \theta )$]
$=\frac{1+\sin \theta }{\cos \theta }\times \frac{1-\sin \theta }{1-\sin \theta }$

$=\frac{1-{\sin }^2\theta }{\cos \theta (1-\sin \theta )}$

$=\frac{{\cos }^2\theta }{\cos \theta (1-\sin \theta )}$

$=\frac{\cos \theta }{1-\sin \theta }$

$\frac{1+\text{sin}({90}^{\circ }-\theta )-{\text{cos}}^2({90}^{\circ }-\theta )}{\text{cos}({90}^{\circ }-\theta )[1+\text{sin}({90}^{\circ }-\theta \left)\right]}$

$=\frac{1+\text{cos}\theta -{\text{sin}}^2\theta }{\text{sin}\theta (1+\text{cos}\theta )}$

$=\frac{(1-{\text{sin}}^2\theta )+\text{cos}\theta }{\text{sin}\theta (1+\text{cos}\theta )}$

$=\frac{{\text{cos}}^2\theta +\text{cos}\theta }{\text{sin}\theta (1+\text{cos}\theta )}$

$=\frac{\text{cos}\theta \left(\text{cos}\theta +1\right)}{\text{sin}\theta \left(1+\text{cos}\theta \right)}=\text{cot}\theta$

$\sqrt{{\text{sec}}^2\theta +{\text{cosec}}^2\theta }=\sqrt{1+{\text{tan}}^2\theta +1+{\text{cot}}^2\theta }$

$=\sqrt{{\text{tan}}^2\theta +2+{\text{cot}}^2\theta }$

$=\sqrt{{\text{tan}}^2\theta +2\left(\text{tan}\theta \right)\left(\text{cot}\theta \right)+{\text{cot}}^2\theta }\dots \dots (\text{tan}\theta \text{cot}\theta =1)$

$=\sqrt{(\text{tan}\theta +\text{cot}\theta {)}^2}$

$=\text{tan}\theta +\text{cot}\theta$

$=\frac{\text{sin}\theta }{\text{cos}\theta }+\frac{\text{cos}\theta }{\text{sin}\theta }$

$=\frac{{\text{sin}}^2\theta +{\text{cos}}^2\theta }{\text{sin}\theta \text{cos}\theta }$

$=\frac{1}{sin\theta cos\theta }$

$=\text{cosec}\theta \text{sec}\theta$