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

Answer:

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

Answer:

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}

Answer:

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})

Answer:

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

Answer:

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 

Answer:

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}

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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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