Question

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Answer 1

Question :

If none of x, y, z is an odd multiple of π/2 and if sin (y + z – x), sin (z + x – y), sin (x + y – z) are in AP, then prove that tanx, tany and tanz are also in AP.

Explaination :

Given that, sin (y + z – x), sin (z + x – y) and sin (x + y – z) are in AP.

➠ sin (z + x – y) – sin (y + z – x) = sin (x +

y – z) – sin (z + x – y)

➠ z cos z sin (x – y) = 2 cosx sinx (y – z)

➠ cosz [sinx cosy – cosxsiny] = cosx

[sinycosz – cosysinz]

Dividing by cosx, cosy and cosz, we get :

➠ sinx/cosx – siny/cosy = siny/cosy –

sinz/cosz

➠ tanx – tany = tany – tanz

➠ tanx + tanz = tany + tany

➠ tanx + tanz = 2 tany

Therefore, tanx, tany and tanz are in AP.

Henceforth, proved!

Answer 2

Answer:

Given that, sin (y + z – x), sin (z + x – y) and sin (x + y – z) are in AP.

➠ sin (z + x – y) – sin (y + z – x) = sin (x +  

y – z) – sin (z + x – y)  

➠ z cos z sin (x – y) = 2 cosx sinx (y – z)  

➠ cosz [sinx cosy – cosxsiny] = cosx  

[sinycosz – cosysinz]  

Dividing by cosx, cosy and cosz, we get :  

➠ sinx/cosx – siny/cosy = siny/cosy –

sinz/cosz  

➠ tanx – tany = tany – tanz  

➠ tanx + tanz = tany + tany  

➠ tanx + tanz = 2 tany

Therefore, tanx, tany and tanz are in AP.

Henceforth, proved!

Step-by-step explanation:

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