To do:We have to simplify \( \left(1+\tan ^{2} \theta\right)(1-\sin \theta)(1+\sin \theta) \).Solution:We know that,$\sin^2 \theta+\cos ^{2} \theta=1$.....(i)$\sec^2 \theta-\tan ^{2} \theta=1$.......(ii)$\sec \theta \times \cos \theta=1$.......(iii)Therefore,$\left(1+\tan ^{2} \theta\right)(1-\sin \theta)(1+\sin \theta)=\left(1+\tan ^{2} \theta\right)(1^2-\sin^2 \theta)$          [Since $(a-b)(a+b)=a^2-b^2$]$=\left(\sec ^{2} \theta\right)(\cos^2 \theta)$           [From (i) and (ii)]$=(\sec \theta\times cos \theta)^2$                  $=1^2$                          [From (iii)]$=1$ Therefore,\( \left(1+\tan ^{2} \theta\right)(1-\sin \theta)(1+\sin \theta)=1 \).

Given:\( \sqrt{3} \tan \theta=1 \)To do:We have to find the value of \( \sin ^{2}-\cos ^{2} \theta \).Solution:  $\sqrt{3} \tan \theta=1$       $\Rightarrow \tan \theta=\frac{1}{\sqrt3}$$\Rightarrow \tan \theta=\tan 30^{\circ}$$\Rightarrow \theta=30^{\circ}$Therefore,$\sin ^{2} \theta-\cos ^{2} \theta=\sin ^{2} 30^{\circ}-\cos ^{2} 30^{\circ}$$=(\frac{1}{2})^{2}-(\frac{\sqrt{3}}{2})^{2}$$=\frac{1}{4}-\frac{3}{4}$$=\frac{1-3}{4}$$=\frac{-2}{4}$$=\frac{-1}{2}$The value of \( \sin ^{2}-\cos ^{2} \theta \) is $\frac{-1}{2}$.

To do:We have to prove that \( 1+\frac{\cot ^{2} \alpha}{1+\operatorname{cosec} \alpha}=\operatorname{cosec} \alpha \).Solution:We know that, $\sin^2 A+\cos^2 A=1$$\operatorname{cosec}^2 A-\cot^2 A=1$$\sec^2 A-\tan^2 A=1$$\cot A=\frac{\cos A}{\sin A}$$\tan A=\frac{\sin A}{\cos A}$$\operatorname{cosec} A=\frac{1}{\sin A}$$\sec A=\frac{1}{\cos A}$Therefore, $1+\frac{\cot ^{2} \alpha}{1+\operatorname{cosec} \alpha}=1+\frac{\frac{\cos ^{2} \alpha}{\sin ^{2} \alpha}}{1+\frac{1}{\sin \alpha}}$$=1+\frac{\frac{\cos ^{2} \alpha}{\sin ^{2} \alpha}}{\frac{\sin \alpha+1}{\sin \alpha}}$$=1+\frac{\cos ^{2} \alpha}{\sin ^{2} \alpha} \times \frac{\sin \alpha}{\sin \alpha+1}$$=1+\frac{\cos ^{2} \alpha}{\sin \alpha(\sin \alpha+1)}$$=\frac{\sin \alpha(\sin \alpha+1)+\cos ^{2} \alpha}{\sin \alpha(\sin \alpha+1)}$$=\frac{\sin^2 \alpha+\sin \alpha+\cos ^{2} \alpha}{\sin \alpha(\sin \alpha+1)}$$=\frac{1+\sin \alpha}{\sin \alpha(\sin \alpha+1)}$$=\frac{1}{\sin \alpha}$$=\operatorname{cosec} \alpha$Hence proved.      Read More

To do:We have to prove that \( (\sqrt{3}+1)(3-\cot 30^{\circ})=\tan^{3} 60^{\circ}-2\sin 60^{\circ} \).Solution:  We know that,$\cot 30^{\circ}=\sqrt3$$\tan 60^{\circ}=\sqrt3$$\sin 60^{\circ}=\frac{\sqrt3}{2}$Let us consider LHS,$(\sqrt{3}+1)(3-\cot 30^{\circ})=(\sqrt{3}+1)(3-\sqrt3)$$=(\sqrt{3})(3)-(\sqrt3)^2+1(3)-1(\sqrt3)$$=3\sqrt3-3+3-\sqrt3$$=2\sqrt3$     Let us consider RHS,$\tan^{3} 60^{\circ}-2\sin 60^{\circ}=(\sqrt3)^3-2(\frac{\sqrt3}{2})$$=3\sqrt3-\sqrt3$$=2\sqrt3$LHS $=$ RHSHence proved. 

Given:\( (\sin \alpha+\cos \alpha)(\tan \alpha+\cot \alpha)=\sec \alpha+\operatorname{cosec} \alpha \) To do:We have to prove that \( (\sin \alpha+\cos \alpha)(\tan \alpha+\cot \alpha)=\sec \alpha+\operatorname{cosec} \alpha \).Solution:We know that, $\tan \theta =\frac{\sin \theta}{\cos \theta}$$\cot \theta=\frac{\cos \theta}{\sin \theta}$$\sin ^{2} \theta+\cos ^{2} \theta=1$$\sec \theta =\frac{1}{\cos \theta}$$\operatorname{cosec} \theta=\frac{1}{\sin \theta}$Therefore, LHS $=(\sin \alpha+\cos \alpha)(\tan \alpha+\cot \alpha)$$=(\sin \alpha+\cos \alpha)(\frac{\sin \alpha}{\cos \alpha}+\frac{\cos \alpha}{\sin \alpha})$$=(\sin \alpha+\cos \alpha)(\frac{\sin^2 \alpha+\cos^2 \alpha}{\sin \alpha \cos \alpha})$$=(\sin \alpha+\cos \alpha) \frac{1}{(\sin \alpha \cos \alpha)}$$=\frac{1}{\cos \alpha}+\frac{1}{\sin \alpha}$$=\sec \alpha+\operatorname{cosec} \alpha$$=$ RHSHence proved.Read More

Given:\( \tan \mathrm{A}=\frac{3}{4} \)To do:We have to prove that \( \sin \mathrm{A} \cos \mathrm{A}=\frac{12}{25} \).Solution:Let, in a triangle $ABC$ right-angled at $B$, $\tan\ A=\frac{3}{4}$.We know that, In a right-angled triangle $ABC$ with right angle at $B$, By Pythagoras theorem, $AC^2=AB^2+BC^2$By trigonometric ratios definitions, $sin\ \theta=\frac{Opposite}{Hypotenuse}=\frac{BC}{AC}$$cos\ \theta=\frac{Adjacent}{Hypotenuse}=\frac{AB}{AC}$$sec\ \theta=\frac{Hypotenuse}{Adjacent}=\frac{AC}{AB}$$tan\ \theta=\frac{Opposite}{Adjacent}=\frac{BC}{AB}$Here, $AC^2=AB^2+BC^2$$\Rightarrow (AC)^2=(4)^2+(3)^2$$\Rightarrow AC^2=16+9$$\Rightarrow AC=\sqrt{25}=5$Therefore, $\sin\ A=\frac{BC}{AC}$$=\frac{3}{5}$$\cos\ A=\frac{AB}{AC}$$=\frac{4}{5}$Therefore, $\sin \mathrm{A} \cos \mathrm{A}=\frac{3}{5} \times \frac{4}{5}$$=\frac{3\times4}{5\times5}$$=\frac{12}{25}$Hence proved. Read More

To do:We have to prove that \( \frac{\tan A}{1+\sec A}-\frac{\tan A}{1-\sec A}=2 \operatorname{cosec} A \).Solution:We know that, $\sin^2 A+\cos^2 A=1$$\operatorname{cosec}^2 A-\cot^2 A=1$$\sec^2 A-\tan^2 A=1$$\cot A=\frac{\cos A}{\sin A}$$\tan A=\frac{\sin A}{\cos A}$$\operatorname{cosec} A=\frac{1}{\sin A}$$\sec A=\frac{1}{\cos A}$Therefore, $\frac{\tan \mathrm{A}}{1+\sec \mathrm{A}}-\frac{\tan \mathrm{A}}{1-\sec \mathrm{A}}=\frac{\tan \mathrm{A}(1-\sec \mathrm{A}-1-\sec \mathrm{A})}{(1+\sec \mathrm{A})(1-\sec \mathrm{A})}$$=\frac{\tan \mathrm{A}(-2 \sec \mathrm{A})}{\left(1-\sec ^{2} \mathrm{~A}\right)}$$=\frac{2 \tan \mathrm{A}\sec \mathrm{A}}{\left(\sec ^{2} \mathrm{~A}-1\right)}$$=\frac{2 \tan \mathrm{A} \cdot \sec \mathrm{A}}{\tan ^{2} \mathrm{~A}}$$=\frac{2 \sec \mathrm{A}}{\tan \mathrm{A}}$$=2\times\frac{1}{\cos A}\times\frac{\cos A}{\sin A}$$=\frac{2}{\sin \mathrm{A}}$$=2 \operatorname{cosec} \mathrm{A}$Hence proved.    Read More

Given:\( \frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}=2 \operatorname{cosec} \theta \)To do:We have to prove that \( \frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}=2 \operatorname{cosec} \theta \)Solution:We know that, $(a+b)^2=a^2+2ab+b^2$$\sin ^{2} \theta+\cos ^{2} \theta=1$Therefore, LHS $=\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}$$=\frac{\sin ^{2} \theta+(1+\cos \theta)^{2}}{\sin \theta(1+\cos \theta)}$$=\frac{\sin ^{2} \theta+1+\cos ^{2} \theta+2 \cos \theta}{\sin \theta(1+\cos \theta)}$$=\frac{1+1+2 \cos \theta}{\sin \theta(1+\cos \theta)}$$=\frac{2(1+\cos \theta)}{\sin \theta(1+\cos \theta)}$$=\frac{2}{\sin \theta}$$=2 \operatorname{cosec} \theta$$=$ RHSHence proved.Read More