Question

if none of A,B and (A+B) is an odd multiple of π/2 then Tan(A-B)=tanA-tanB/1+tanA+tanB​

Answer 1

Step-by-step explanation:

here u go hope it helps you man

Answer 2

Concept: An area of standardised mathematics called trigonometry studies how lengths, heights, and angles relate to one another. To answer trigonometric problems, trigonometry has its own set of trigonometric angles, equations, and functions.

Trigonometry has always been a component of advanced mathematics, which has applications in practically every subject, including physics, astronomy, architecture, geography, research, and everyday life. Although trigonometry isn't directly employed in daily life, it is used in the majority of the household equipment. It can be used for navigation, computing, programming, imaging in medicine, measuring the heights of mountains and structures, etc.

Given: None of A, B and (A+B) is an odd multiple of $\frac{\pi }{2}$

To find: Prove that

$tan (A-B) = \frac{tan A - tan B}{1 + tan Atan B}$

Solution:

Before deriving this formula, first we need to know some basic additional formulas of trigonometry

$sin (A + B) = sinAcosB + cosAsinB$

$sin(A - B) = sinAcosB - cosAsinB$

$cos(A + B) = cosAcosB - sinAsinB$

$cos(A - B) = cosAcosB + sinAsinB$

As we know, the formula can be derived from earlier addition formulae.

$tan (A - B) = \frac{sin (A - B)}{cos (A - B)}$

When the formulas for sin(A-B) and cos(A-B) are combined, the outcome is

$tan(A-B) = \frac{sinAcosB - cosAsinB}{cosAcosB + sinAsinB}$

Now that we have divided each right-hand side word by cosAcosB, we have:

$tan(A-B) = \frac{\frac{sinAcosB}{cosAcosB} - \frac{cosAsinB}{cosAcosB} }{\frac{cosAcosB}{cosAcosB} + \frac{sinAsinB}{cosAcosB} }$

Leaving away the components that are common,

$tan (A-B) = \frac{\frac{sinA}{cosA} - \frac{sinB}{cosB} }{1+\frac{sinAsinB}{cosAcosB} }$

As we know $tan = \frac{sin}{cos}$

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

Hence, the formula is derived.

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