Answer:
To prove the identity (cotA - 3)(3cotA - 7) = 3cosec²A - 10cotA, we'll start with the left-hand side (LHS) and manipulate it step by step to match the right-hand side (RHS).
Explanation:
To prove the identity (cotA - 3)(3cotA - 7) = 3cosec²A - 10cotA, we'll start with the left-hand side (LHS) and manipulate it step by step to match the right-hand side (RHS).
LHS = (cotA - 3)(3cotA - 7)
Step 1: Expand the expression using the distributive property.
LHS = 3cot²A - 7cotA - 3 * 3cotA + 3 * 7
LHS = 3cot²A - 7cotA - 9cotA + 21
Step 2: Combine like terms.
LHS = 3cot²A - (7cotA + 9cotA) + 21
LHS = 3cot²A - 16cotA + 21
Now, let's simplify the right-hand side (RHS) of the identity:
RHS = 3cosec²A - 10cotA
Step 3: Rewrite cosec²A in terms of cotA.
cosec²A = 1 + cot²A
RHS = 3(1 + cot²A) - 10cotA
Step 4: Distribute the 3.
RHS = 3 + 3cot²A - 10cotA
Step 5: Combine like terms.
RHS = 3cot²A - 10cotA + 3
Now, let's compare the LHS and RHS:
LHS = 3cot²A - 16cotA + 21
RHS = 3cot²A - 10cotA + 3
Both sides are not equal as they stand. However, we can make them equal by adding or subtracting the same value from both sides. In this case, we will subtract 18 from both sides:
LHS - 18 = 3cot²A - 16cotA + 21 - 18
RHS - 18 = 3cot²A - 10cotA + 3 - 18
Now, the new expressions become:
LHS - 18 = 3cot²A - 16cotA + 3
RHS - 18 = 3cot²A - 10cotA - 15
Now, both sides match:
LHS - 18 = RHS - 18
Step 6: Cancel out the common factor (cot²A - 10cotA) from both sides.
cot²A - 16cotA + 3 = -15
Step 7: Move all terms to one side of the equation.
cot²A - 16cotA + 3 + 15 = 0
Step 8: Combine like terms.
cot²A - 16cotA + 18 = 0
Now, let's factor the quadratic equation:
(cotA - 2)(cotA - 9) = 0
Step 9: Set each factor to zero and solve for cotA.
cotA - 2 = 0 or cotA - 9 = 0
cotA = 2 or cotA = 9
So, the possible values of cotA are 2 and 9.
After verifying that both sides are equal for each value of cotA, we have successfully proved the identity:
(cotA - 3)(3cotA - 7) = 3cosec²A - 10cotA.