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

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

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

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

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

$= 2$

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

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

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

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

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

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

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

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

LHS $(cosec \ A - sin \ A)(sec \ A - cos \ A)$

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

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

$= sin \ A . cos \ A$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$= cosec \ A + sec \ A$

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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 $sec \ A + cosec \ A = n$, show that $n(m^2-1)= 2m$

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

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

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

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

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

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.

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