\displaystyle \text{Question 1: } \frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}=\sin A+\cos A \hspace{1.0cm} [\text{ICSE }2003]
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}
\displaystyle =\frac{\cos^2 A}{\cos A-\sin A}-\frac{\sin^2 A}{\cos A-\sin A}
\displaystyle =\frac{\cos^2 A-\sin^2 A}{\cos A-\sin A}
\displaystyle =\frac{(\cos A-\sin A)(\cos A+\sin A)}{\cos A-\sin A}
\displaystyle =\cos A+\sin A=\text{RHS}.
\displaystyle \therefore \text{ Hence proved.}
\displaystyle \\

\displaystyle \text{Question 2: } \frac{\cos^3 A+\sin^3 A}{\cos A+\sin A}+\frac{\cos^3 A-\sin^3 A}{\cos A-\sin A}=2
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\cos^3 A+\sin^3 A}{\cos A+\sin A}+\frac{\cos^3 A-\sin^3 A}{\cos A-\sin A}
\displaystyle =\frac{\cos^3 A+\sin^3 A}{\cos A+\sin A}\times\frac{\cos A-\sin A}{\cos A-\sin A}+\frac{\cos^3 A-\sin^3 A}{\cos A-\sin A}\times\frac{\cos A+\sin A}{\cos A+\sin A}
\displaystyle =\frac{\cos^4 A+\sin^3 A\cos A-\cos^3 A\sin A-\sin^4 A}{\cos^2 A-\sin^2 A}+\frac{\cos^4 A-\sin^3 A\cos A+\cos^3 A\sin A-\sin^4 A}{\cos^2 A-\sin^2 A}
\displaystyle =\frac{(1-\sin A\cos A)(\cos^2 A-\sin^2 A)}{\cos^2 A-\sin^2 A}+\frac{(1+\sin A\cos A)(\cos^2 A-\sin^2 A)}{\cos^2 A-\sin^2 A}
\displaystyle =1-\sin A\cos A+1+\sin A\cos A
\displaystyle =2=\text{RHS}.
\displaystyle \therefore \text{Hence proved.}
\displaystyle \\

\displaystyle \text{Question 3: } \sec A\cdot\mathrm{cosec}\,A+1=\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}
\displaystyle \text{Answer:}
\displaystyle \text{RHS}=\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}
\displaystyle =\frac{\sin^2 A}{\cos A(\sin A-\cos A)}+\frac{\cos^2 A}{\sin A(\cos A-\sin A)}
\displaystyle =\frac{\sin^3 A-\cos^3 A}{\sin A\cos A(\sin A-\cos A)}
\displaystyle =\frac{(\sin A-\cos A)(\sin^2 A+\sin A\cos A+\cos^2 A)}{\sin A\cos A(\sin A-\cos A)}
\displaystyle =\frac{\sin^2 A+\cos^2 A+\sin A\cos A}{\sin A\cos A}
\displaystyle =\frac{1+\sin A\cos A}{\sin A\cos A}
\displaystyle =\frac{1}{\sin A\cos A}+1
\displaystyle =\sec A\cdot\mathrm{cosec}\,A+1=\text{LHS}.
\displaystyle \therefore \text{Hence proved.}
\displaystyle \\

\displaystyle \text{Question 4: } \left(\tan A+\frac{1}{\cos A}\right)^2+\left(\tan A-\frac{1}{\cos A}\right)^2=2\left(\frac{1+\sin^2 A}{1-\sin^2 A}\right)
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\left(\tan A+\frac{1}{\cos A}\right)^2+\left(\tan A-\frac{1}{\cos A}\right)^2
\displaystyle =\frac{(\sin A+1)^2}{\cos^2 A}+\frac{(\sin A-1)^2}{\cos^2 A}
\displaystyle =\frac{\sin^2 A+1+2\sin A+\sin^2 A+1-2\sin A}{\cos^2 A}
\displaystyle =\frac{2\sin^2 A+2}{\cos^2 A}
\displaystyle =\frac{2(1+\sin^2 A)}{1-\sin^2 A}
\displaystyle =2\left(\frac{1+\sin^2 A}{1-\sin^2 A}\right)=\text{RHS}.
\displaystyle \therefore \text{Hence proved.}
\displaystyle \\

\displaystyle \text{Question 5: } 2\sin^2 A+\cos^4 A=1+\sin^4 A
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=2\sin^2 A+\cos^4 A
\displaystyle =2\sin^2 A+(1-\sin^2 A)^2
\displaystyle =2\sin^2 A+1+\sin^4 A-2\sin^2 A
\displaystyle =1+\sin^4 A=\text{RHS}.
\displaystyle \therefore \text{Hence proved.}
\displaystyle \\

\displaystyle \text{Question 6: } \frac{\sin A-\sin B}{\cos A+\cos B}+\frac{\cos A-\cos B}{\sin A+\sin B}=0
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\sin A-\sin B}{\cos A+\cos B}+\frac{\cos A-\cos B}{\sin A+\sin B}
\displaystyle =\frac{(\sin A-\sin B)(\sin A+\sin B)+(\cos A-\cos B)(\cos A+\cos B)}{(\cos A+\cos B)(\sin A+\sin B)}
\displaystyle =\frac{\sin^2 A-\sin^2 B+\cos^2 A-\cos^2 B}{(\cos A+\cos B)(\sin A+\sin B)}
\displaystyle =\frac{\sin^2 A-(1-\cos^2 B)+(1-\sin^2 A)-\cos^2 B}{(\cos A+\cos B)(\sin A+\sin B)}
\displaystyle =0=\text{RHS}.
\displaystyle \therefore \text{Hence proved.}
\displaystyle \\

\displaystyle \text{Question 7: } (\mathrm{cosec}\,A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=(\mathrm{cosec}\,A-\sin A)(\sec A-\cos A)
\displaystyle =\left(\frac{1}{\sin A}-\sin A\right)\left(\frac{1}{\cos A}-\cos A\right)
\displaystyle =\frac{1-\sin^2 A}{\sin A}\cdot\frac{1-\cos^2 A}{\cos A}
\displaystyle =\frac{\cos^2 A\sin^2 A}{\sin A\cos A}
\displaystyle =\sin A\cos A
\displaystyle \text{RHS}=\frac{1}{\tan A+\cot A}=\frac{1}{\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}}
\displaystyle =\frac{1}{\frac{\sin^2 A+\cos^2 A}{\sin A\cos A}}=\sin A\cos A
\displaystyle \therefore \text{LHS}=\text{RHS}. \text{ Hence proved.}
\displaystyle \\

\displaystyle \text{Question 8: } (1+\tan A\tan B)^2+(\tan A-\tan B)^2=\sec^2 A\sec^2 B
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=(1+\tan A\tan B)^2+(\tan A-\tan B)^2
\displaystyle =\frac{(\cos A\cos B+\sin A\sin B)^2}{\cos^2 A\cos^2 B}+\frac{(\sin A\cos B-\sin B\cos A)^2}{\cos^2 A\cos^2 B}
\displaystyle =\frac{\cos^2 A\cos^2 B+\sin^2 A\sin^2 B+2\sin A\sin B\cos A\cos B}{\cos^2 A\cos^2 B}+\frac{\sin^2 A\cos^2 B+\sin^2 B\cos^2 A-2\sin A\sin B\cos A\cos B}{\cos^2 A\cos^2 B}
\displaystyle =\frac{\cos^2 A\cos^2 B+\sin^2 A\sin^2 B+\sin^2 A\cos^2 B+\sin^2 B\cos^2 A}{\cos^2 A\cos^2 B}
\displaystyle =\frac{\cos^2 A(\cos^2 B+\sin^2 B)+\sin^2 A(\cos^2 B+\sin^2 B)}{\cos^2 A\cos^2 B}
\displaystyle =\frac{\cos^2 A+\sin^2 A}{\cos^2 A\cos^2 B}
\displaystyle =\frac{1}{\cos^2 A\cos^2 B}
\displaystyle =\sec^2 A\sec^2 B=\text{RHS}.
\displaystyle \therefore \text{Hence proved.}
\displaystyle \\

\displaystyle \text{Question 9: } \frac{1}{\cos A+\sin A-1}+\frac{1}{\cos A+\sin A+1}=\mathrm{cosec}\,A+\sec A
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{1}{\cos A+\sin A-1}+\frac{1}{\cos A+\sin A+1}
\displaystyle =\frac{\cos A+\sin A+1+\cos A+\sin A-1}{(\cos A+\sin A)^2-1}
\displaystyle =\frac{2(\cos A+\sin A)}{\cos^2 A+\sin^2 A+2\sin A\cos A-1}
\displaystyle =\frac{2(\cos A+\sin A)}{2\sin A\cos A}
\displaystyle =\frac{\cos A+\sin A}{\sin A\cos A}
\displaystyle =\frac{1}{\sin A}+\frac{1}{\cos A}
\displaystyle =\mathrm{cosec}\,A+\sec A=\text{RHS}.
\displaystyle \therefore \text{Hence proved.}
\displaystyle \\

\displaystyle \text{Question 10: If }x\cos A+y\sin A=m\text{ and }x\sin A-y\cos A=n, \\ \text{ then prove that }x^2+y^2=m^2+n^2
\displaystyle \text{Answer:}
\displaystyle \text{Given: }x\cos A+y\sin A=m\text{ and }x\sin A-y\cos A=n
\displaystyle \text{Squaring both sides, we get}
\displaystyle m^2=x^2\cos^2 A+y^2\sin^2 A+2xy\sin A\cos A \qquad \ldots\ldots\ldots\text{(i)}
\displaystyle n^2=x^2\sin^2 A+y^2\cos^2 A-2xy\sin A\cos A \qquad \ldots\ldots\ldots\text{(ii)}
\displaystyle \text{Adding (i) and (ii), we get}
\displaystyle m^2+n^2=x^2\cos^2 A+y^2\sin^2 A+x^2\sin^2 A+y^2\cos^2 A
\displaystyle =x^2(\cos^2 A+\sin^2 A)+y^2(\sin^2 A+\cos^2 A)
\displaystyle =x^2+y^2
\displaystyle \therefore x^2+y^2=m^2+n^2. \text{ Hence proved.}
\displaystyle \\

\displaystyle \text{Question 11: If }m=a\sec A+b\tan A\text{ and }n=a\tan A+b\sec A, \\ \text{ then prove that }m^2-n^2=a^2-b^2
\displaystyle \text{Answer:}
\displaystyle \text{Given: }m=a\sec A+b\tan A\text{ and }n=a\tan A+b\sec A
\displaystyle \text{Squaring both sides, we get}
\displaystyle m^2=a^2\sec^2 A+b^2\tan^2 A+2ab\sec A\tan A \qquad \ldots\ldots\ldots\text{(i)}
\displaystyle n^2=a^2\tan^2 A+b^2\sec^2 A+2ab\sec A\tan A \qquad \ldots\ldots\ldots\text{(ii)}
\displaystyle \text{Subtracting (ii) from (i), we get}
\displaystyle m^2-n^2=a^2\sec^2 A+b^2\tan^2 A-a^2\tan^2 A-b^2\sec^2 A
\displaystyle =a^2(\sec^2 A-\tan^2 A)+b^2(\tan^2 A-\sec^2 A)
\displaystyle =a^2(1)-b^2(1)
\displaystyle =a^2-b^2
\displaystyle \therefore m^2-n^2=a^2-b^2. \text{ Hence proved.}
\displaystyle \\

\displaystyle \text{Question 12: If }x=r\sin A\cos B,\;y=r\sin A\sin B\text{ and }z=r\cos A, \\ \text{ then prove that }x^2+y^2+z^2=r^2
\displaystyle \text{Answer:}
\displaystyle \text{Given: }x=r\sin A\cos B,\;y=r\sin A\sin B\text{ and }z=r\cos A
\displaystyle \text{Squaring all and adding, we get}
\displaystyle x^2+y^2+z^2=r^2\sin^2 A\cos^2 B+r^2\sin^2 A\sin^2 B+r^2\cos^2 A
\displaystyle =r^2(\sin^2 A\cos^2 B+\sin^2 A\sin^2 B+\cos^2 A)
\displaystyle =r^2\left(\sin^2 A(\cos^2 B+\sin^2 B)+\cos^2 A\right)
\displaystyle =r^2(\sin^2 A+\cos^2 A)
\displaystyle =r^2
\displaystyle \therefore x^2+y^2+z^2=r^2. \text{ Hence proved.}
\displaystyle \\

\displaystyle \text{Question 13: If }\sin A+\cos A=m\text{ and }\sec A+\mathrm{cosec}\,A=n, \\ \text{ show that }n(m^2-1)=2m
\displaystyle \text{Answer:}
\displaystyle \text{Given: }\sin A+\cos A=m\text{ and }\sec A+\mathrm{cosec}\,A=n
\displaystyle \text{Therefore }m^2=(\sin A+\cos A)^2=1+2\sin A\cos A
\displaystyle \Rightarrow m^2-1=2\sin A\cos A
\displaystyle \Rightarrow n(m^2-1)=(\sec A+\mathrm{cosec}\,A)\cdot2\sin A\cos A
\displaystyle =\left(\frac{1}{\cos A}+\frac{1}{\sin A}\right)\cdot2\sin A\cos A
\displaystyle =\frac{\sin A+\cos A}{\sin A\cos A}\cdot2\sin A\cos A
\displaystyle =2(\sin A+\cos A)
\displaystyle =2m
\displaystyle \therefore n(m^2-1)=2m. \text{ Hence proved.}
\displaystyle \\

\displaystyle \text{Question 14: If }x=r\cos A\cos B,\;y=r\cos A\sin B\text{ and }z=r\sin A,\text{ show that } \\ x^2+y^2+z^2=r^2
\displaystyle \text{Answer:}
\displaystyle \text{Given: }x=r\cos A\cos B,\;y=r\cos A\sin B\text{ and }z=r\sin A
\displaystyle \text{Squaring all the three equations and adding, we get}
\displaystyle x^2+y^2+z^2=r^2\cos^2 A\cos^2 B+r^2\cos^2 A\sin^2 B+r^2\sin^2 A
\displaystyle =r^2(\cos^2 A\cos^2 B+\cos^2 A\sin^2 B+\sin^2 A)
\displaystyle =r^2\left(\cos^2 A(\cos^2 B+\sin^2 B)+\sin^2 A\right)
\displaystyle =r^2(\cos^2 A+\sin^2 A)
\displaystyle =r^2
\displaystyle \therefore x^2+y^2+z^2=r^2. \text{ Hence proved.}
\displaystyle \\

\displaystyle \text{Question 15: If }\frac{\cos A}{\cos B}=m\text{ and }\frac{\cos A}{\sin B}=n,\text{ show that }(m^2+n^2)\cos^2 B=n^2
\displaystyle \text{Answer:}
\displaystyle \text{Given: }\frac{\cos A}{\cos B}=m\Rightarrow m^2=\frac{\cos^2 A}{\cos^2 B}
\displaystyle \frac{\cos A}{\sin B}=n\Rightarrow n^2=\frac{\cos^2 A}{\sin^2 B}
\displaystyle m^2+n^2=\frac{\cos^2 A}{\cos^2 B}+\frac{\cos^2 A}{\sin^2 B}
\displaystyle =\cos^2 A\left(\frac{1}{\cos^2 B}+\frac{1}{\sin^2 B}\right)
\displaystyle =\frac{\cos^2 A(\sin^2 B+\cos^2 B)}{\cos^2 B\sin^2 B}
\displaystyle =\frac{\cos^2 A}{\cos^2 B\sin^2 B}
\displaystyle =\frac{n^2}{\cos^2 B}
\displaystyle \Rightarrow (m^2+n^2)\cos^2 B=n^2
\displaystyle \therefore \text{Hence proved.}
\displaystyle \\

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