Prove that tanA tan(90 - A) = cotA cot(90 - A) = sinA sec(90 - A) = - Maths - Trigonometric Functions - 13499875

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tan(90 + x), tan(90 + A), tan(90 + theta), Identity for tan(90 + x)

How to prove that 1/cotA (1-cotA) + 1/tanA (1- tanA) = 1+secA.cosecA - Quora

tan(90 + x), tan(90 + A), tan(90 + theta), Identity for tan(90 + x)

Prove the following identities: sin A cos A (tanA+cotA)=1

How to prove tan⁡𝑥+tan⁡(90°−𝑥) = sec⁡𝑥csc⁡𝑥?

What is the formula of tan(90 - A)? - Quora

$\\=\frac{(1+\cot A+\tan A)(\sin A-\cos A)}{sec^3 A - \csc^3 A}\\\\\\ =\frac{\left (1+\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} \right )(\sin A-\cos A)}{\frac{1}{\cos^3A}-\frac{1}{\sin^3A}}\\\\\\ =\frac{\left (\frac{\sin A.\cos A+\sin^2A+\cos^2A}{\sin A .\cos A} \right )(\sin A-\cos A)}{\frac{\sin^3A-\cos^3A}{\sin^3A.\cos^3A}}\\\\\\ =\frac{\left (\sin A.\cos A+\sin^2A+\cos^2A \right )(\sin A-\cos A)}{\sin A .\cos A}\times\frac{\sin^3A.\cos^3A}{\sin^3A-\cos^3A}\\\\\\ (\text{By the identity: } (a-b)(a^2+b^2+ab)=a^3-b^3)\\\\\\ =\frac{\sin^3A-\cos^3A}{\sin A .\cos A}\times\frac{\sin^3A.\cos^3A}{\sin^3A-\cos^3A}\\\\\\ =\sin^2A.\cos^2a\\\\=RHS$

Show that:(1+cotA+tanA)(sinA-cosA) / sec^3 A - cosec^3 A Introduction to Trigonometry-Maths-Class-10

Tan (90- Ө) = Cot Ө and Cot (90- Ө) = Tan Ө

prove that cot(90-A)/tanA +cosec(90-A)sinA /tan(90-A)=sec (square )A

If A + B = 90°, then (tan A tan B + tan A cot B)/(sin A sec B) - sin^2 B/cos^2 A is equal to A. cot^2A - Sarthaks eConnect

prove that tan(90-A)+cot(90-A)/sec(90-A)=cosec(90-A)