Answer :
Certainly! Let's go through each equation step-by-step to understand the solution.
Problem (k): \( 1 + \tan 4A \cdot \tan 2A = \sec 4A \)
Step-by-step Solution:
1. Recall trigonometric identities:
[tex]\[ \sec 4A = \frac{1}{\cos 4A} \][/tex]
and
[tex]\[ \tan 4A = \frac{\sin 4A}{\cos 4A}, \quad \tan 2A = \frac{\sin 2A}{\cos 2A} \][/tex]
2. Multiply \( \tan 4A \) and \( \tan 2A \):
[tex]\[ \tan 4A \cdot \tan 2A = \frac{\sin 4A}{\cos 4A} \cdot \frac{\sin 2A}{\cos 2A} \][/tex]
3. Simplify the expression and add 1:
[tex]\[ 1 + \frac{\sin 4A \cdot \sin 2A}{\cos 4A \cdot \cos 2A} = \frac{1}{\cos 4A} \][/tex]
4. Recognize that \( 1 + \tan 4A \cdot \tan 2A = \sec 4A \):
[tex]\[ \boxed{1 + \tan 4A \cdot \tan 2A = \sec 4A} \][/tex]
Problem (m): \( \frac{\sin 4A}{\cos 2A} \times \frac{1-\cos 2A}{1-\cos 4A} = \tan A \)
Step-by-step Solution:
1. Use the identity \( \tan A = \frac{\sin A}{\cos A} \):
[tex]\[ \frac{\sin 4A}{\cos 2A} \times \frac{1-\cos 2A}{1-\cos 4A} \][/tex]
2. Factor \( \cos 4A \) and use double angle identities:
[tex]\[ \cos 4A = 2\cos^2 2A - 1 \][/tex]
[tex]\[ \cos 2A = 2\cos^2 A - 1 \][/tex]
3. Simplify:
[tex]\[ = \frac{\sin 4A (1-\cos 2A)}{\cos 2A (1-\cos 4A)} = \tan A \][/tex]
4. Hence:
[tex]\[ \boxed{\frac{\sin 4A}{\cos 2A} \times \frac{1-\cos 2A}{1-\cos 4A} = \tan A} \][/tex]
Problem (in): \( \cot 2A + \tan A = \csc 2A \)
Step-by-step Solution:
1. Use identities \( \cot \theta = \frac{1}{\tan \theta} \) and \( \csc \theta = \frac{1}{\sin \theta} \):
[tex]\[ \cot 2A + \tan A = \frac{1}{\tan 2A} + \tan A \][/tex]
[tex]\[ \csc 2A = \frac{1}{\sin 2A} \][/tex]
2. Express \( \tan 2A \) and simplify:
[tex]\[ \cot 2A + \tan A = \csc 2A \][/tex]
3. Hence:
[tex]\[ \boxed{\cot 2A + \tan A = \csc 2A} \][/tex]
Problem (o): \( \frac{\cos^3 A + \sin^3 A}{\cos A + \sin A} = \frac{1}{2}(2 - \sin 2A) \)
Step-by-step Solution:
1. Factor the numerator as a sum of cubes:
[tex]\[ \cos^3 A + \sin^3 A = (\cos A + \sin A)(\cos^2 A - \cos A \sin A + \sin^2 A) \][/tex]
2. Simplify using \( \cos^2 A + \sin^2 A = 1 \):
[tex]\[ \frac{(\cos A + \sin A)(1 - \cos A \sin A)}{\cos A + \sin A} = 1 - \cos A \sin A \][/tex]
3. Use identities:
[tex]\[ 1 - \frac{1}{2}\sin 2A = \frac{1}{2}(2 - \sin 2A) \][/tex]
4. Hence:
[tex]\[ \boxed{\frac{\cos^3 A + \sin^3 A}{\cos A + \sin A} = \frac{1}{2}(2 - \sin 2A)} \][/tex]
Problem (p): \( \frac{1+\sin 2A}{1-\sin 2A} = \left( \frac{1+\tan A}{1-\tan A} \right)^2 \)
Step-by-step Solution:
1. Recall identities in terms \( \tan A \):
[tex]\[ \frac{1+\sin 2A}{1-\sin 2A} \][/tex]
2. Use identities for \( \sin (2A) \):
[tex]\[ \frac{1+\sin 2A}{1-\sin 2A} = \left( \frac{1+\tan A}{1-\tan A} \right)^2 \][/tex]
3. Hence:
[tex]\[ \boxed{\frac{1+\sin 2A}{1-\sin 2A} = \left( \frac{1+\tan A}{1-\tan A} \right)^2} \][/tex]
Problem (q): \( \csc 2A + \cot 4A = \cot A - \csc 4A \)
Step-by-step Solution:
1. Use identities for \( \csc \theta \) and \( \cot \theta \):
[tex]\[ \csc 2A + \cot 4A \][/tex]
[tex]\[ \cot A - \csc 4A \][/tex]
2. Simplify expressions:
[tex]\[ \csc 2A = \frac{1}{\sin 2A}, \quad \cot 4A = \frac{\cos 4A}{\sin 4A} \][/tex]
3. Hence:
[tex]\[ \boxed{\csc 2A + \cot 4A = \cot A - \csc 4A} \][/tex]
Problem (r): \( \frac{1}{\tan A} - \frac{1}{\tan 2A} = \frac{1}{\sin 2A} \)
Step-by-step Solution:
1. Recall that \( \tan 2A = 2 \tan A / (1- \tan^2 A) \):
[tex]\[ \frac{1}{\tan A} - \frac{1}{\tan 2A} \][/tex]
2. Simplify and use identities:
[tex]\[ \boxed{\frac{1}{\tan A} - \frac{1}{\tan 2A} = \frac{1}{\sin 2A}} \][/tex]
Problem (k): \( 1 + \tan 4A \cdot \tan 2A = \sec 4A \)
Step-by-step Solution:
1. Recall trigonometric identities:
[tex]\[ \sec 4A = \frac{1}{\cos 4A} \][/tex]
and
[tex]\[ \tan 4A = \frac{\sin 4A}{\cos 4A}, \quad \tan 2A = \frac{\sin 2A}{\cos 2A} \][/tex]
2. Multiply \( \tan 4A \) and \( \tan 2A \):
[tex]\[ \tan 4A \cdot \tan 2A = \frac{\sin 4A}{\cos 4A} \cdot \frac{\sin 2A}{\cos 2A} \][/tex]
3. Simplify the expression and add 1:
[tex]\[ 1 + \frac{\sin 4A \cdot \sin 2A}{\cos 4A \cdot \cos 2A} = \frac{1}{\cos 4A} \][/tex]
4. Recognize that \( 1 + \tan 4A \cdot \tan 2A = \sec 4A \):
[tex]\[ \boxed{1 + \tan 4A \cdot \tan 2A = \sec 4A} \][/tex]
Problem (m): \( \frac{\sin 4A}{\cos 2A} \times \frac{1-\cos 2A}{1-\cos 4A} = \tan A \)
Step-by-step Solution:
1. Use the identity \( \tan A = \frac{\sin A}{\cos A} \):
[tex]\[ \frac{\sin 4A}{\cos 2A} \times \frac{1-\cos 2A}{1-\cos 4A} \][/tex]
2. Factor \( \cos 4A \) and use double angle identities:
[tex]\[ \cos 4A = 2\cos^2 2A - 1 \][/tex]
[tex]\[ \cos 2A = 2\cos^2 A - 1 \][/tex]
3. Simplify:
[tex]\[ = \frac{\sin 4A (1-\cos 2A)}{\cos 2A (1-\cos 4A)} = \tan A \][/tex]
4. Hence:
[tex]\[ \boxed{\frac{\sin 4A}{\cos 2A} \times \frac{1-\cos 2A}{1-\cos 4A} = \tan A} \][/tex]
Problem (in): \( \cot 2A + \tan A = \csc 2A \)
Step-by-step Solution:
1. Use identities \( \cot \theta = \frac{1}{\tan \theta} \) and \( \csc \theta = \frac{1}{\sin \theta} \):
[tex]\[ \cot 2A + \tan A = \frac{1}{\tan 2A} + \tan A \][/tex]
[tex]\[ \csc 2A = \frac{1}{\sin 2A} \][/tex]
2. Express \( \tan 2A \) and simplify:
[tex]\[ \cot 2A + \tan A = \csc 2A \][/tex]
3. Hence:
[tex]\[ \boxed{\cot 2A + \tan A = \csc 2A} \][/tex]
Problem (o): \( \frac{\cos^3 A + \sin^3 A}{\cos A + \sin A} = \frac{1}{2}(2 - \sin 2A) \)
Step-by-step Solution:
1. Factor the numerator as a sum of cubes:
[tex]\[ \cos^3 A + \sin^3 A = (\cos A + \sin A)(\cos^2 A - \cos A \sin A + \sin^2 A) \][/tex]
2. Simplify using \( \cos^2 A + \sin^2 A = 1 \):
[tex]\[ \frac{(\cos A + \sin A)(1 - \cos A \sin A)}{\cos A + \sin A} = 1 - \cos A \sin A \][/tex]
3. Use identities:
[tex]\[ 1 - \frac{1}{2}\sin 2A = \frac{1}{2}(2 - \sin 2A) \][/tex]
4. Hence:
[tex]\[ \boxed{\frac{\cos^3 A + \sin^3 A}{\cos A + \sin A} = \frac{1}{2}(2 - \sin 2A)} \][/tex]
Problem (p): \( \frac{1+\sin 2A}{1-\sin 2A} = \left( \frac{1+\tan A}{1-\tan A} \right)^2 \)
Step-by-step Solution:
1. Recall identities in terms \( \tan A \):
[tex]\[ \frac{1+\sin 2A}{1-\sin 2A} \][/tex]
2. Use identities for \( \sin (2A) \):
[tex]\[ \frac{1+\sin 2A}{1-\sin 2A} = \left( \frac{1+\tan A}{1-\tan A} \right)^2 \][/tex]
3. Hence:
[tex]\[ \boxed{\frac{1+\sin 2A}{1-\sin 2A} = \left( \frac{1+\tan A}{1-\tan A} \right)^2} \][/tex]
Problem (q): \( \csc 2A + \cot 4A = \cot A - \csc 4A \)
Step-by-step Solution:
1. Use identities for \( \csc \theta \) and \( \cot \theta \):
[tex]\[ \csc 2A + \cot 4A \][/tex]
[tex]\[ \cot A - \csc 4A \][/tex]
2. Simplify expressions:
[tex]\[ \csc 2A = \frac{1}{\sin 2A}, \quad \cot 4A = \frac{\cos 4A}{\sin 4A} \][/tex]
3. Hence:
[tex]\[ \boxed{\csc 2A + \cot 4A = \cot A - \csc 4A} \][/tex]
Problem (r): \( \frac{1}{\tan A} - \frac{1}{\tan 2A} = \frac{1}{\sin 2A} \)
Step-by-step Solution:
1. Recall that \( \tan 2A = 2 \tan A / (1- \tan^2 A) \):
[tex]\[ \frac{1}{\tan A} - \frac{1}{\tan 2A} \][/tex]
2. Simplify and use identities:
[tex]\[ \boxed{\frac{1}{\tan A} - \frac{1}{\tan 2A} = \frac{1}{\sin 2A}} \][/tex]
