$\displaystyle \text{Question 1: } \frac{\cos A}{1 - \tan A}+\frac{\sin A}{1 - \cot A} = \sin A + \cos A \hspace{1.0cm} [2003]$

$\displaystyle \text{LHS } = \frac{\cos A}{1 - \tan A}+\frac{ \sin A}{1 - \cot A}$

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

$\displaystyle = \frac{(\cos A - \sin A)(\cos A + \sin A)}{\cos A - \sin A}$

$\displaystyle = \sin A + \cos A =$ RHS. Hence proved.

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$\displaystyle \text{Question 2: } \frac{\cos^3 A + \sin^3 A}{\cos A+ \sin A} +$ $\displaystyle \frac{\cos^3 A - \sin^3 A}{\cos A - \sin A} = 2$

$\displaystyle \text{LHS } = \frac{\cos^3 A + \sin^3 A}{\cos A+ \sin A} +$ $\displaystyle \frac{\cos^3 A - \sin^3 A}{\cos A - \sin A}$

$\displaystyle = \frac{\cos^3 A + \sin^3 A}{\cos A+ \sin A} \times \frac{\cos A - \sin A}{\cos A - \sin A} + \frac{\cos^3 A - \sin^3 A}{\cos A - \sin A} \times \frac{\cos A + \sin A }{\cos A + \sin A}$

$\displaystyle = \frac{\cos^4 A + \sin^3 A .\cos A - \cos^3 A .\sin A - \sin^4 A}{\cos^2 A - \sin^2 A} +$ $\displaystyle \frac{\cos^4 A - \sin^3 A .\cos A + \cos^3 A .\sin A - \sin^4 A}{\cos^2 A - \sin^2 A}$

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

$\displaystyle = 1 - \sin A .\cos A + 1 + \sin A .\cos A$

$\displaystyle = 2$

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$\displaystyle \text{Question 3: } \sec A . \mathrm{cosec} A + 1 = \frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A}$

$\displaystyle \text{RHS } = \frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A}$

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

$\displaystyle = \frac{\sin^3 A - \cos^3 A}{\cos A (\sin A - \cos A) \sin A}$

$\displaystyle = \frac{(\sin A - \cos A)^3 + 3 \sin A .\cos A(\sin A - \cos A)}{\cos A (\sin A - \cos A) \sin A}$

$\displaystyle = \frac{\sin^2 A + \cos^2 - 2 \sin A .\cos A + 3 \sin A .\cos A}{\sin A .\cos A}$

$\displaystyle = \frac{1+ \sin A .\cos A}{\sin A .\cos A}$

$\displaystyle = \sec A. \mathrm{cosec} A + 1 =$ LHS. Hence proved.

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$\displaystyle \text{Question 4: } (\tan A + \frac{1}{\cos A} )^2 + (\tan A + \frac{1}{\cos A} )^2 = 2 (\frac{1+\sin^2 A}{1-\sin^2 A})$

$\displaystyle \text{LHS } = (\tan A + \frac{1}{\cos A} )^2 + (\tan A + \frac{1}{\cos A} )^2 = 2$

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

$\displaystyle = \frac{\sin^2 A + 1 + 2 \sin A + \sin^2 A + 1 - 2 \sin A}{\cos^2 A}$

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

$\displaystyle = 2 (\frac{1+ \sin^2 A}{1-\sin^2 A})$ = RHS. Hence proved.

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$\displaystyle \text{Question 5: } 2 \sin^2 A + \cos^4 A = 1 + \sin^4 A$

$\displaystyle \text{LHS } = 2 \sin^2 A + \cos^4 A$

$\displaystyle = 2 \sin^2 A + (1-\sin^2 A)^2$

$\displaystyle = 2 \sin^2 A + 1 + \sin^4 A - 2 \sin^2 A$

$\displaystyle = 1+ \sin^4 A =$ RHS. Hence proved.

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$\displaystyle \text{Question 6: } \frac{\sin A - \sin B}{\cos A + \cos B}+\frac{\cos A - \cos B}{\sin A + \sin B} =0$

$\displaystyle \text{LHS } = \frac{\sin A - \sin B}{\cos A + \cos B}+\frac{\cos A - \cos B}{\sin A + \sin B}$

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

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

$\displaystyle = 0 =$ RHS. Hence proved.

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$\displaystyle \text{Question 7: } (\mathrm{cosec} A - \sin A)(\sec A - \cos A) = \frac{1}{\tan A + \cot A}$

$\displaystyle \text{LHS } (\mathrm{cosec} A - \sin A)(sec A - \cos A)$

$\displaystyle = \frac{1 - \sin^2 A}{\sin A} . \frac{1 - \cos^2 A}{\cos A} = \frac{1 - \sin^2 A - \cos^2A + \sin^2 A.\cos^2 A}{\sin A. \cos A} = \sin A . \cos A$

$\displaystyle \text{RHS } = \frac{1}{\tan A + \cot A} = \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}} = \sin A. \cos A$

Therefore LHS = RHS. Hence proved.

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$\displaystyle \text{Question 8: } (1 + \tan A. \tan B)^2+(\tan A - \tan B)^2 = \sec^2 A . \sec^2 B$

$\displaystyle \text{LHS } = (1 + \tan A. \tan B)^2+(\tan A - \tan B)^2$

$\displaystyle = \frac{\cos A. \cos B + \sin A. \sin B)^2}{\cos^2 A.\sin^2 B} + \frac{\sin A. \cos B - \sin B. \cos A)^2}{\cos^2 A.\sin^2 B}$

$\displaystyle = \frac{\cos^2 A.\cos^2 B+\sin^2 A. \sin^2 B + 2\cos A. \cos B.\sin A. \sin B}{\cos^2 A.\sin^2 B} + \frac{\sin^2 A. \cos^2 B + \sin^2 B. \cos^2 A - 2\sin A. \cos B.\sin B. \cos A }{\cos^2 A.\sin^2 B}$

$\displaystyle = \frac{\cos^2 A.\cos^2 B+\sin^2 A. \sin^2 B + \sin^2 A. \cos^2 B + \sin^2 B. \cos^2 A }{\cos^2 A.\sin^2 B}$

$\displaystyle = \frac{\cos^2 A (\cos^2 B + \sin^2 B) + \sin^2 A(\cos^2 B + \sin^2 B)}{\cos^2 A.\sin^2 B}$

$\displaystyle = \frac{\cos^2 A + \sin^2 A}{\cos^2 A.\sin^2 B}$

$\displaystyle = \frac{1}{\cos^2 A.\sin^2 B}$

$\displaystyle = \sec^2 A . \sec^2 B =$ RHS. Hence proved.

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$\displaystyle \text{Question 9: } \frac{1}{\cos A + \sin A - 1}+ \frac{1}{\cos A + \sin A + 1} = \mathrm{cosec} A + \sec A$

$\displaystyle \text{LHS } = \frac{1}{\cos A + \sin A - 1}+ \frac{1}{\cos A + \sin A + 1}$

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

$\displaystyle = \frac{2(\cos A + \sin A)}{1 + 2 \cos A. \sin A - 1}$

$\displaystyle = \frac{\cos A + \sin A }{\cos A. \sin A}$

$\displaystyle = \mathrm{cosec} A + \sec A$

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$\displaystyle \text{Question 10: If } x \cos A + y \sin A = m$ and $\displaystyle x \sin A -y \cos A = n$ , then prove that $\displaystyle x^2 + y^2 = m^2 +n^2$

$\displaystyle \text{Given: } x \cos A + y \sin A = m$ and $\displaystyle x \sin A -y \cos A = n$

Squaring both sides we get:

$\displaystyle m^2 = x^2 \cos^2 A + y^2 \sin^2 A + 2 x^2 \cos^2 A . y^2 \sin^2 A$ … … … (i)

$\displaystyle n^2 = x^2 \sin^2 A + y^2 \cos^2 A - 2 x^2 \cos^2 A . y^2 \sin^2 A$ … … … (ii)

Adding (i) and (ii), we get

$\displaystyle m^2 + n^2 = x^2 \cos^2 A + y^2 \sin^2 A + x^2 \sin^2 A + y^2 \cos^2 A$

$\displaystyle \Rightarrow m^2 + n^2 = x^2 (\cos^2 A + \sin^2 A ) + y^2 (\sin^2 A + \cos^2 A)$

$\displaystyle m^2 + n^2 = x^2 + y^2$ . Hence proved.

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$\displaystyle \text{Question 11: If } m = a \sec A + b \tan A$ and $\displaystyle n = a \tan A + b \sec A$ , then prove that $\displaystyle m^2 -n^2 = a^2 - b^2$

$\displaystyle \text{Given: } m = a \sec A + b \tan A$ and $\displaystyle n = a\tan A + b\sec A$

Squaring both sides:

$\displaystyle m^2 = a^2 \sec^2 A + b^2 \tan^2 A + 2a^2 b^2 \sec^2 A . \tan^2 A$ … … … (i)

$\displaystyle n^2 = a^2\tan^2 A + b^2\sec^2 A + 2a^2 b^2 \sec^2 A . \tan^2 A$ … … … (ii)

Subtracting (ii) from (i) we get

$\displaystyle m^2 - n^2 = a^2 \sec^2 A + b^2 \tan^2 A + 2a^2 b^2 \sec^2 A . \tan^2 A - a^2\tan^2 A - b^2\sec^2 A - 2a^2 b^2 \sec^2 A . \tan^2 A$

$\displaystyle m^2 - n^2 = a^2 \sec^2 A + b^2 \tan^2 A - a^2\tan^2 A - b^2\sec^2 A$

$\displaystyle m^2 - n^2 = a^2 (\sec^2 A - \tan^2 A) + b^2 (\tan^2 A -\sec^2 A)$

$\displaystyle m^2 - n^2 = a^2 (\frac{1-\sin^2 A}{\cos^2 A}) + b^2 (\frac{\sin^2 A - 1}{\cos^2 A})$

$\displaystyle m^2 - n^2 = a^2 (\frac{\cos^2 A}{\cos^2 A}) + b^2 (\frac{-\cos^2 A}{\cos^2 A})$

$\displaystyle m^2 - n^2 = a^2 - b^2$

Hence proved.

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$\displaystyle \text{Question 12: If } x = r \sin A .\cos B$ , $\displaystyle y = r \sin A .\sin B$ and $\displaystyle z = r \cos A$ , then prove that $\displaystyle x^2 +y^2 +z^2 = r^2$

$\displaystyle \text{Given: } x = r \sin A .\cos B$ , $\displaystyle y = r \sin A .\sin B$ and $\displaystyle z = r \cos A$

Squaring all and adding, we get:

$\displaystyle x^2 + y^2 + z^2 = r^2 \sin^2 A .\cos^2 B + r^2 \sin^2 A .\sin^2 B + r^2 \cos^2 A$

$\displaystyle x^2 + y^2 + z^2 = r^2 (\sin^2 A .\cos^2 B + \sin^2 A .\sin^2 B + \cos^2 A)$

$\displaystyle x^2 + y^2 + z^2 = r^2 (\sin^2 A (\cos^2 B + \sin^2 B) + \cos^2 A)$

$\displaystyle x^2 + y^2 + z^2 = r^2 (\sin^2 A + \cos^2 A)$

$\displaystyle x^2 + y^2 + z^2 = r^2$

Hence proved.

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$\displaystyle \text{Question 13: If } \sin A + \cos A = m$ and $\displaystyle \sec A + \mathrm{cosec} A = n$ , show that $\displaystyle n(m^2-1)= 2m$

$\displaystyle \text{Given: } \sin A + \cos A = m$ and $\displaystyle \sec A + \mathrm{cosec} A = n$

$\displaystyle \text{Therefore } m^2 = 1 + 2.\sin A. \cos A \Rightarrow (m^2 -1) = 2 \sin A.\cos A$

$\displaystyle \Rightarrow n(m^2 -1) = (\sec A + \mathrm{cosec} A). 2 \sin A.\cos A$

$\displaystyle = (\frac{\sin A + \cos A}{\sin A.\cos A}) .2 \sin A.\cos A$

$\displaystyle = 2(\sin A + \cos A) = 2m$

Hence proved.

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$\displaystyle \text{Question 14: If } x = r \cos A . \cos B$ , $\displaystyle y = r \cos A . \sin B$ and $\displaystyle z = r \sin A$ , show that $\displaystyle x^2 + y^2 +z^2 = r^2$

$\displaystyle \text{Given: } x = r \cos A . \cos B$ , $\displaystyle y = r \cos A . \sin B$ and $\displaystyle z = r \sin A$

Squaring all the three equations and adding, we get

$\displaystyle x^2 + y^2 + z^2 = r^2 \cos^2 A . \cos^2 B + r^2 \cos^2 A . \sin^2 B + r^2 \sin^2 A$

$\displaystyle \Rightarrow x^2 + y^2 + z^2= r^2 (\cos^2 A . \cos^2 B + \cos^2 A . \sin^2 B + \sin^2 A)$

$\displaystyle \Rightarrow x^2 + y^2 + z^2 = r^2 (\cos^2 A (\cos^2 B + \sin^2 B) + \sin^2 A)$

$\displaystyle \Rightarrow x^2 + y^2 + z^2 = r^2 (\cos^2 A + \sin^2 A)$

$\displaystyle \Rightarrow x^2 + y^2 + z^2 = r^2$

Hence proved.

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$\displaystyle \text{Question 15: If } \frac{\cos A}{\cos B} = m \text{ and } \frac{\cos A}{\sin B} = n \text{, show that } (m^2 +n^2) \cos^2 B = n^2$

$\displaystyle \text{Given: } \frac{\cos A}{\cos B} = m \Rightarrow m^2 = \frac{\cos^2 A}{\cos^2 B}$

$\displaystyle \frac{\cos A}{\sin B} = n \Rightarrow n^2 = \frac{\cos^2 A}{\cos^2 B}$

$\displaystyle m^2 + n^2 = \cos ^2 A (\frac{1}{\cos^2 B} + \frac{1}{\sin^2 B})$

$\displaystyle = \frac{\cos^2 A}{\cos^2 B. \sin^2 B}$

$\displaystyle = \frac{n^2}{\cos^2 B}$

$\displaystyle \Rightarrow (m^2 + n^2). \cos^2 B = n^2$

Hence proved.