tanA - sin A /sin^2A= tan A / 1 + cosA
Please prove this
Did you mean:
(tanA - sinA)/tan^2 A = tanA/(1+cosA)
the way you typed it, it is not true
LS = (tanA-sinA)/sin^2 A
= (sinA/cosA - sinA)/sin^2 A
= ( (sinA - sinAcosA)/cosA) /sin^2 A
= (sinA(1 - cosA)/cosA)/sin^2 A
= (1 - cosA)/(sinAcosA)
RS = (sinA/cosA) / (1 + cosA)
= (sinA/cosA) / (1 + cosA) * (1-cosA)/(1-cosA)
= (sinA/cosA)(1-cosA)/(1 - cos^2 A)
= (sinA/cosA)(1-cosA)/sin^2 A
= (1-cosA)/(sinAcosA)
= LS
Thanks reiny sir
Excellent
Excellent
Why did the triangle go to a therapist? Because it couldn't solve its trigonometric problems on its own! Now, let's prove this equation step by step.
Starting with the left side:
tan A - sin A / sin^2 A
We can rewrite tan A as sin A / cos A, so the equation becomes:
sin A / cos A - sin A / sin^2 A
To simplify, let's find a common denominator by multiplying the second term by cos^2 A and the first term by sin^2 A:
(sin A * cos^2 A) / (cos A * sin^2 A) - sin A / sin^2 A
Now, let's combine the fractions:
(sin A * cos^2 A - sin A) / (cos A * sin^2 A)
Factoring out the common factor of sin A:
sin A * (cos^2 A - 1) / (cos A * sin^2 A)
Remember that cos^2 A - 1 is equal to -sin^2 A, so we can rewrite the equation as:
sin A * (-sin^2 A) / (cos A * sin^2 A)
Canceling out the sin^2 A terms:
- sin A / cos A
And we know that - sin A is equal to sin A, so the left side simplifies to:
sin A / cos A, which is equal to tan A.
Now, let's focus on the right side of the equation:
tan A / (1 + cos A)
Since tan A is sin A / cos A, the equation becomes:
(sin A / cos A) / (1 + cos A)
To divide two fractions, we multiply the first fraction by the reciprocal of the second fraction:
(sin A / cos A) * (1 / (1 + cos A))
Simplifying further:
sin A / (cos A * (1 + cos A))
We can multiply the denominator by (1 - cos A) / (1 - cos A) to try and find a common factor:
sin A / ((cos A * (1 + cos A)) * (1 - cos A) / (1 - cos A))
Expanding the denominator:
sin A / (cos A - cos^2 A + cos A - cos^2 A)
Simplifying:
sin A / (-2cos^2 A + 2cos A)
Factoring out 2cos A:
sin A / (2cos A (cos A - 1))
We know that cos^2 A - 1 is equal to -sin^2 A, so we can rewrite the equation as:
sin A / (2cos A (-sin^2 A))
Canceling out the sin^2 A terms:
- sin A / (2cos A)
And we know that - sin A is equal to sin A, so the right side simplifies to:
sin A / (2cos A)
Now, comparing the left side (tan A - sin A / sin^2 A) and the right side (sin A / (1 + cos A)), we can see that they simplify to the same expression, which is tan A.
So, we have now proved that:
tan A - sin A / sin^2 A = sin A / (1 + cos A)
Keep those trigonometry problems coming, and I'll be here to clown around with solutions for you!
To prove the given equation:
Start with the left-hand side (LHS) of the equation:
LHS = (tan A - sin A) / sin^2 A
Now, let's simplify the LHS step by step:
Step 1: Distribute the denominator sin^2 A:
LHS = (tan A - sin A) / (sin A * sin A)
Step 2: Bring the common factor sin A outside the numerator:
LHS = sin A * (tan A - sin A) / (sin A * sin A)
Step 3: Cancel out the common factor sin A in the numerator and denominator:
LHS = (tan A - sin A) / sin A
Moving to the right-hand side (RHS) of the equation:
RHS = tan A / (1 + cos A)
Now, let's simplify the RHS step by step:
Step 1: Bring the common factor cos A outside the denominator:
RHS = tan A / (1 * (1 + cos A))
Step 2: Cancel out the common factor 1 in the numerator and denominator:
RHS = tan A / (1 + cos A)
Comparing the simplified LHS and RHS, we can conclude that they are the same:
LHS = RHS
Hence, we have proven that tan A - sin A / sin^2 A = tan A / (1 + cos A).