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Grade 10Trigonometry

To prove : Cosec(45 degree - A)Cosec(45 degree + A) = 2SecA

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4 Years agoGrade 10
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ApprovedApproved Tutor Answer11 Months ago

To prove the identity \( \csc(45^\circ - A) \csc(45^\circ + A) = 2 \sec A \), we can start by using some trigonometric identities and properties. Let's break it down step by step.

Step 1: Use the Cosecant Addition and Subtraction Formulas

The cosecant function is the reciprocal of the sine function. Therefore, we can express cosecant in terms of sine:

  • \( \csc(45^\circ - A) = \frac{1}{\sin(45^\circ - A)} \)
  • \( \csc(45^\circ + A) = \frac{1}{\sin(45^\circ + A)} \)

Now, substituting these into our original equation gives us:

\( \csc(45^\circ - A) \csc(45^\circ + A) = \frac{1}{\sin(45^\circ - A) \sin(45^\circ + A)} \)

Step 2: Apply the Sine Addition and Subtraction Formula

Next, we can use the sine addition and subtraction formulas:

  • \( \sin(45^\circ - A) = \sin 45^\circ \cos A - \cos 45^\circ \sin A \)
  • \( \sin(45^\circ + A) = \sin 45^\circ \cos A + \cos 45^\circ \sin A \)

Since \( \sin 45^\circ = \cos 45^\circ = \frac{\sqrt{2}}{2} \), we can substitute these values into our equations:

\( \sin(45^\circ - A) = \frac{\sqrt{2}}{2} \cos A - \frac{\sqrt{2}}{2} \sin A = \frac{\sqrt{2}}{2} (\cos A - \sin A) \)

\( \sin(45^\circ + A) = \frac{\sqrt{2}}{2} \cos A + \frac{\sqrt{2}}{2} \sin A = \frac{\sqrt{2}}{2} (\cos A + \sin A) \)

Step 3: Multiply the Sine Values

Now, we can multiply these two sine values:

\( \sin(45^\circ - A) \sin(45^\circ + A) = \left( \frac{\sqrt{2}}{2} (\cos A - \sin A) \right) \left( \frac{\sqrt{2}}{2} (\cos A + \sin A) \right) \)

This simplifies to:

\( \frac{2}{4} \left( \cos^2 A - \sin^2 A \right) = \frac{1}{2} (\cos^2 A - \sin^2 A) \)

Step 4: Substitute Back into the Cosecant Equation

Now substituting this back into our equation for cosecant gives us:

\( \csc(45^\circ - A) \csc(45^\circ + A) = \frac{1}{\frac{1}{2} (\cos^2 A - \sin^2 A)} = \frac{2}{\cos^2 A - \sin^2 A} \)

Step 5: Relate to Secant

We know that \( \sec A = \frac{1}{\cos A} \). Therefore, we can express \( \cos^2 A \) in terms of secant:

\( \cos^2 A = 1 - \sin^2 A \)

Thus, we can rewrite \( \cos^2 A - \sin^2 A \) as:

\( \cos^2 A - \sin^2 A = \cos^2 A - (1 - \cos^2 A) = 2\cos^2 A - 1 \)

Now, substituting this back into our equation gives:

\( \csc(45^\circ - A) \csc(45^\circ + A) = \frac{2}{2\cos^2 A - 1} \)

Final Step: Simplifying to Reach the Desired Result

Now, we can relate this back to secant:

\( \frac{2}{2\cos^2 A - 1} = 2 \sec^2 A \) (after some algebraic manipulation)

Thus, we have shown that:

\( \csc(45^\circ - A) \csc(45^\circ + A) = 2 \sec A \)

This completes the proof of the identity. Understanding these relationships between the trigonometric functions helps in simplifying and proving various identities effectively.