Question 1:   $\frac{\cos A}{1 - \tan A}+\frac{sin A}{1 - \cot A}$ $= \sin A + \cos A$   [2003]

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

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

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

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

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

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

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

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

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

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

$= 2$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Question 5:   $2 \sin^2 \ A + \cos^4 \ A = 1 + \sin^4 \ A$

LHS $= 2 \sin^2 A + \cos^4 A$

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

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

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

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

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

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

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

$= 0 =$ RHS. Hence proved.

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

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

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

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

LHS $= (1 + \tan A. \tan B)^2+(\tan A - \tan B)^2$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Given: $x \cos A + y \sin A = m$ and $x \sin A -y \cos A = n$

Squaring both sides we get:

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

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

Adding (i) and (ii), we get

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

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

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

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

Given: $m = a \sec A + b \tan A$ and $n = a\tan A + b\sec A$

Squaring both sides:

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

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

Subtracting (ii) from (i) we get

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

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

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

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

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

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

Hence proved.

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

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

Squaring all and adding, we get:

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

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

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

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

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

Hence proved.

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

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

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

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

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

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

Hence proved.

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

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

Squaring all the three equations and adding, we get

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

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

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

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

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

Hence proved.

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

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

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

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

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

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

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

Hence proved.