Question

prove that (secA-cosA)(cotA+tanA)=tanA.secA​

Answer 1

$\red\bigstar$ Explanation $\red\bigstar$

$\leadsto$ Given:-

LHS:-

(secA-cosA)(cotA+tanA)

($\dfrac{1}{cosA}$ - cosA)($\dfrac{cosA}{sinA}$ + $\dfrac{sinA}{cosA}$)

($\dfrac{(1-cos²A)}{cosA}$)($\dfrac{(sin²A + cos²A)}{sinAcosA}$)

We know that

sin²A + cos²A = 1

Therefore,

sin²A = 1 - cos²A

($\dfrac{(sin²A)}{cosA}$)($\dfrac{1}{sinAcosA}$)

($\dfrac{(sinA)}{cosA}$)($\dfrac{1}{cosA}$)

We know that tanA = $\dfrac{(sinA)}{cosA}$

We also know that secA = $\dfrac{1}{cosA}$

LHS = RHS

Therefore,

(secA - cosA)(cotA + tanA) = tanA.secA

$\leadsto$ Formulas related to Trigonometry:-

tanA = $\dfrac{sinA}{cosA}$

cotA = $\dfrac{cosA}{sinA}$

cotA = $\dfrac{1}{tanA}$

secA = $\dfrac{1}{cosA}$

cosecA = $\dfrac{1}{sinA}$

sin²A + cos²A = 1

1 + tan²A = sec²A

1 + cot²A = cosec²A

sin(A + B) = sinAcosB + cosAsinB

sin(A - B) = sinAcosB - cosAsinB

cos(A + B) = cosAcosB - sinAsinB

cos(A - B) = cosAcosB + sinAsinB

tan(A + B) = $\dfrac{(tanA + tanB)}{1 - tanAtanB}$

tan(A - B) = $\dfrac{(tanA - tanB)}{1 + tanAtanB}$

cot(A + B) = $\dfrac{(cotAcotB - 1)}{(cotA + cotB)}$

cot(A - B) = $\dfrac{(cotAcotB + 1)}{(cotA - cotB)}$

sin2A = 2sinAcosA

cos2A = cos²A - sin²A

tan2A = $\dfrac{2tanA}{1 - tan²A}$

cot2A = $\dfrac{(cot²A - 1)}{2cotA}$

sin3A = 3sinA - 4sin³A

cos3A = 4cos³A - 3cosA

tan3A = $\dfrac{3tanA - tan³A}{1 - 3tan²A}$