If cosec |-sin |=l and sec |- cos |=m, prove that l2m2(l2+m2+3)=1
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Step-by-Step Solution
Step 1: Express l and m in terms of sinθ and cosθ
We are given the expressions for l and m in terms of cosecant, sine, secant, and cosine. We convert cosecant and secant into their reciprocal forms, sinθ1 and cosθ1 respectively. Then, we combine the terms by finding a common denominator and use the identity sin2θ+cos2θ=1 to simplify the numerators.
Step 2: Calculate l2 and m2
Now that we have simplified expressions for l and m, we can find l2 and m2 by squaring both sides of each equation. This will give us the terms needed for the expression we need to prove.
Step 3: Calculate l2m2
We multiply the expressions for l2 and m2. Notice that sin2θ in the denominator of l2 cancels with two powers of sinθ in the numerator of m2, and similarly for cos2θ.
Step 4: Calculate l2+m2
To add l2 and m2, we find a common denominator, which is sin2θcos2θ. This results in cos6θ+sin6θ in the numerator.
Step 5: Simplify l2+m2+3
We add 3 to the expression for l2+m2. We use the algebraic identity a3+b3=(a+b)(a2−ab+b2) where a=sin2θ and b=cos2θ. We also use the identity a2+b2=(a+b)2−2ab to simplify sin4θ+cos4θ. After simplifying, we combine the terms with 3.
Step 6: Substitute into the expression to prove
Finally, we substitute the calculated values of l2m2 and l2+m2+3 into the expression we need to prove. The terms sin2θcos2θ cancel out, leaving us with 1.