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.

\\

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 - \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

\\

Question 3:   sec \ A . \mathrm{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^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.

\\

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.

\\

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.

\\

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.

\\

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

Answer:

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.

\\

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

\\

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

=  \mathrm{cosec} \ A + \sec \ A 

\\

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.

\\

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.

\\

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.

\\

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

Answer:

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.

\\

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.

\\

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.