Prove that
[1+cotA-CosecA] [1+tanA+secA] = 2
[1+cotA-CosecA]*[1+tanA+secA]
= 1 + tanA + secA + cotA + 1 + cosecA - (1/sinA) - (1/cosA) - secAcosecA
= 1 + tanA + secA + cotA + 1 + cosecA - cosecA - secA - secAcosecA
= 2 + tanA + cotA - secAcosecA
= 2 + [(sin^2A + cos^2A)/(sinAcosA)] - secAcosecA
= 2 + 1/(sinAcosA) - secAcosecA
= 2 + secAcosecA - secAcosecA
= 2
1+cot-csc = (sin+cos-1)/sin
1+tan+sec = (sin+cos+1)/cos
multiply them and you have
((sin+cos)^2-1)/(sin*cos)
= (1+2sin*cos-1)/(sin*cos)
= 2
To prove that
[1+cotA-CosecA] [1+tanA+secA] = 2,
we will simplify the left-hand side expression step by step.
Step 1:
Using the distributive property, let's expand the expression:
[1+cotA-CosecA] [1+tanA+secA]
= 1 * (1+tanA+secA) + cotA * (1+tanA+secA) - CosecA * (1+tanA+secA)
Step 2:
Simplifying the terms in each parenthesis, we get:
= 1 + tanA + secA + cotA + tanA * cotA + secA * cotA - cotA * CosecA - CosecA * tanA - CosecA * secA
Step 3:
Let's simplify the trigonometric expressions using their respective definitions:
tanA = sinA / cosA
cotA = cosA / sinA
secA = 1 / cosA
CosecA = 1 / sinA
Substituting these values, we get:
= 1 + sinA / cosA + 1 / cosA + cosA / sinA + (sinA / cosA) * (cosA / sinA) + (1 / cosA) * (cosA / sinA) - (cosA / sinA) * (1 / sinA) - (1 / sinA) * (sinA / cosA) - (1 / sinA) * (1 / cosA)
Step 4:
Simplifying further, we have:
= 1 + sinA / cosA + 1 / cosA + cosA / sinA + 1 + 1 - 1 - 1 - 1
= 3 + sinA / cosA + cosA / sinA
Step 5:
Now, let's simplify the expression sinA / cosA + cosA / sinA.
multiplying both numerator and denominator of sinA / cosA by sinA, we get:
sinA^2 / (cosA*sinA)
multiplying both numerator and denominator of cosA / sinA by cosA, we get:
cosA^2 / (sinA*cosA)
Combining these fractions, we have:
(sinA^2 + cosA^2) / (sinA*cosA)
Using the trigonometric identity, sinA^2 + cosA^2 = 1, we can simplify further:
= 1 / (sinA*cosA)
Step 6:
Substituting the value of 1 / (sinA*cosA) back into our expression:
= 3 + 1 / (sinA*cosA)
Step 7:
Now, let's simplify the expression 1 / (sinA*cosA) further:
Using the trigonometric identity, secA = 1 / cosA and cscA = 1 / sinA:
1 / (sinA*cosA) = secA * cscA
Substituting this back into our expression:
= 3 + secA * cscA
Step 8:
Finally, using the trigonometric identity, secA * cscA = 1, we find:
= 3 + 1
= 4
Therefore, we have shown that:
[1+cotA-CosecA] [1+tanA+secA] = 2+2 = 4
Hence, the original equation is proved.
To prove the given expression:
[1 + cotA - cosecA] [1 + tanA + secA] = 2
We can start by simplifying each term individually and then multiply the simplified expressions:
1 + cotA - cosecA = (sinA/cosA + cosA/sinA) - (1/sinA)
1 + cotA - cosecA = (sinA + cos^2A - 1) / (cosA * sinA)
Similarly,
1 + tanA + secA = (sinA/cosA + sinA/cosA) + (1/cosA)
1 + tanA + secA = (sin^2A + 2cosA + cos^2A) / (cosA * sinA)
Now, let's multiply both expressions:
[(sinA + cos^2A - 1) / (cosA * sinA)] * [(sin^2A + 2cosA + cos^2A) / (cosA * sinA)]
To simplify this expression further, we can cancel out common factors:
(sinA + cos^2A - 1) * (sin^2A + 2cosA + cos^2A) / (cosA * sinA)^2
Expanding the brackets:
sinA * sin^2A + 2cosA * sinA + cos^2A * sinA + cos^2A * sin^2A
+ cos^2A * sin^2A + 2cosA * sinA + cosA * sin^2A - sinA - cos^2A + 1
Combining like terms:
sin^3A + 2cosA * sinA + sinA * cos^2A + cos^2A * sin^2A + cos^2A * sin^2A
+ 2cosA * sinA + cosA * sin^2A - sinA - cos^2A + 1
Now, simplify the expression further:
sin^3A + 2cosA * sinA + cosA * sin^2A + 2cosA * sinA + cosA * sin^2A - sinA - cos^2A + 1
Rearranging the terms:
sin^3A + 2cos^2A * sinA + 2cosA * sinA + 2cosA * sinA + cosA * sin^2A - sinA - cos^2A + 1
Combining like terms:
sin^3A + 4cosA * sinA + cosA * sin^2A - sinA - cos^2A + 1
This expression does not simplify directly to 2. Hence, the given expression is not equal to 2.