If A, B and C are interior angles of a triangle ABC then show that,
[tex]\mathsf{tan^2 \left ( \dfrac{B+C}{2} \right ) = cosec^2 \: \dfrac{A}{2} -1}[/tex]
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[tex]\mathsf{tan^2 \left ( \dfrac{B+C}{2} \right ) = cosec^2 \: \dfrac{A}{2} -1}[/tex]
[tex]\Large \boxed {\mathsf {\red {WARNING }}}[/tex]
[tex]\mathsf{\blue {\dag {Don't \: Spam}}}[/tex]
[tex]\mathsf{\pink {\dag {Want \: Answers\: from\: Moderators}}}[/tex]
Answers & Comments
Verified answer
Solution
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Casually, we take the three sides as a, b, and c, and the opposite angles as A, B, and C.
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[tex]\text{$A+B+C=\pi$ $\dots$[Eqn.1]}[/tex]
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Equation 1 follows the below.
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[tex]\text{$\tan^{2}\bigg(\dfrac{B+C}{2}\bigg)$}[/tex]
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[tex]\text{$=\tan^{2}\bigg(\dfrac{\pi-A}{2}\bigg)$}[/tex]
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[tex]\text{$=\tan^{2}\bigg(\dfrac{\pi}{2}-\dfrac{A}{2}\bigg)$}[/tex]
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[tex]\text{$=\cot^{2}\dfrac{A}{2}$}[/tex]
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In a right triangle, the equation applies.
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[tex]\text{$\sin^{2}\theta+\cos^{2}\theta=1$}[/tex]
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Then,
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[tex]\text{$\sin^{2}\theta+\cos^{2}\theta=1$}[/tex]
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[tex]\text{$\dfrac{\sin^{2}\theta}{\sin^{2}\theta}+\dfrac{\cos^{2}\theta}{\sin^{2}\theta}=\csc^{2}\theta$}[/tex]
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[tex]\text{$\therefore 1+\cot^{2}\theta=\csc^{2}\theta$ $\dots$[Eqn.2]}[/tex]
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Equation 2 follows the below.
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[tex]\text{$\cot^{2}\dfrac{A}{2}$}[/tex]
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[tex]\text{$=\csc^{2}\dfrac{A}{2}-1$}[/tex]
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If we integrate all the calculations⋯
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[tex]\text{$\tan^{2}\bigg(\dfrac{B+C}{2}\bigg)$}[/tex]
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[tex]\text{$=\cot^{2}\dfrac{A}{2}$}[/tex]
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[tex]\text{$=\csc^{2}\dfrac{A}{2}-1$}[/tex]
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The LHS equals the RHS. Now, we find that the equation is always true.
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Learn More
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How to Convert
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The problem is about the trigonometric functions.
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Now, refer to the image.
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★ The First to Remember
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1st - 2nd - 3rd - 4th
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A(All) - S(Sine) - T(Tangent) - C(Cosine)
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[Add Sugar To Coffee, for mnemonic purposes.]
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It is because the functions are defined on a circle.
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1st Step
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When an angle [tex]\text{$\dfrac{n\pi}{2}\pm\theta$}[/tex] is odd in n, change the function.
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When an angle [tex]\text{$\dfrac{n\pi}{2}\pm\theta$}[/tex] is even in n, keep the function.
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2nd Step
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Consider θ as an acute angle.
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Note that we keep the sign of the original function where the angles lie in which quadrant.
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I hope you got it clear. Thanks for reading!
Answer:
I think up answer is correct