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For any Δ ABC show that (c2 – a2 + b2) tan A = (a2 – b2 + c2) tan B = (b2 – c2 + a2) tan C
For any Δ ABC show that (c2 – a2 + b2) tan A = (a2 – b2 + c2) tan B = (b2 – c2 + a2) tan C
CBSE | Class 11 | Exercise 10.2 | Maths | RD Sharma | Sine and Cosine Formulae and their Applications
For any Δ ABC show that (c2 – a2 + b2) tan A = (a2 – b2 + c2) tan B = (b2 – c2 + a2) tan C

Considering L.H.S :

\[({{c}^{2}}~\text{ }{{a}^{2}}~+\text{ }{{b}^{2}}),\text{ }({{a}^{2}}~\text{ }{{b}^{2}}~+\text{ }{{c}^{2}}),\text{ }({{b}^{2}}~\text{ }{{c}^{2}}~+\text{ }{{a}^{2}})\]

Using sine rule in \[\Delta \text{ }ABC\]

RD Sharma Solutions for Class 11 Maths Chapter 10 – Sine and Cosine Formulae and their Applications image - 38

\[({{c}^{2}}~\text{ }{{a}^{2}}~+\text{ }{{b}^{2}}),\text{ }({{a}^{2}}~\text{ }{{b}^{2}}~+\text{ }{{c}^{2}})\text{ }and~({{b}^{2}}~\text{ }{{c}^{2}}~+\text{ }{{a}^{2}}),\] are present in L.H.S. which can be obtained from cosine formulae.

Using cosine formula:

\[Cos\text{ }A\text{ }=\text{ }({{b}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{a}^{2}})/2bc\]

\[2bc\text{ }cos\text{ }A\text{ }=\text{ }({{b}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{a}^{2}})\]

Multiplying tan A both sides we get,

\[2bc\text{ }cos\text{ }A\text{ }tan\text{ }A\text{ }=\text{ }({{b}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{a}^{2}})\text{ }tan\text{ }A\]

\[2bc\text{ }cos\text{ }A\text{ }\left( sin\text{ }A/cos\text{ }A \right)\]

\[=\text{ }({{b}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{a}^{2}})\text{ }tan\text{ }A\]

\[2bc\text{ }sin\text{ }A\text{ }=\text{ }({{b}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{a}^{2}})\text{ }tan\text{ }A\text{ }\ldots \text{ }\left( i \right)\]

Or,

\[Cos\text{ }B\text{ }=\text{ }({{a}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{b}^{2}})/2ac\]

\[2ac\text{ }cos\text{ }B\text{ }=\text{ }({{a}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{b}^{2}})\]

Multiplying tan B both the sides,

\[2ac\text{ }cos\text{ }B\text{ }tan\text{ }B\text{ }=\text{ }({{a}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{b}^{2}})\text{ }tan\text{ }B\]

\[2ac\text{ }cos\text{ }B\text{ }\left( sin\text{ }B/cos\text{ }B \right)\]

\[=\text{ }({{a}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{b}^{2}})\text{ }tan\text{ }B\]

Or,

\[2ac\text{ }sin\text{ }B\text{ }=\text{ }({{a}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{b}^{2}})\text{ }tan\text{ }B\text{ }\ldots \text{ }\left( ii \right)\]

\[Cos\text{ }C\text{ }=\text{ }({{a}^{2}}~+\text{ }{{b}^{2}}~\text{ }{{c}^{2}})/2ab\]

Or,

\[2ab\text{ }cos\text{ }C\text{ }=\text{ }({{a}^{2}}~+\text{ }{{b}^{2}}~\text{ }{{c}^{2}})\]

Multiplying  tan C both sides,

\[2ab\text{ }cos\text{ }C\text{ }tan\text{ }C\text{ }=\text{ }({{a}^{2}}~+\text{ }{{b}^{2}}~\text{ }{{c}^{2}})\text{ }tan\text{ }C\]

\[2ab\text{ }cos\text{ }C\text{ }\left( sin\text{ }C/cos\text{ }C \right)\text{ }=\text{ }({{a}^{2}}~+\text{ }{{b}^{2}}~\text{ }{{c}^{2}})\text{ }tan\text{ }C\]

Or,

\[2ab\text{ }sin\text{ }C\text{ }=\text{ }({{a}^{2}}~+\text{ }{{b}^{2}}~\text{ }{{c}^{2}})\text{ }tan\text{ }C\text{ }\ldots \text{ }\left( iii \right)\]

Using sine formula,

RD Sharma Solutions for Class 11 Maths Chapter 10 – Sine and Cosine Formulae and their Applications image - 39

Multiplying a,b,c in each expression,

RD Sharma Solutions for Class 11 Maths Chapter 10 – Sine and Cosine Formulae and their Applications image - 40

\[bc\text{ }sin\text{ }A\text{ }=\text{ }ac\text{ }sin\text{ }B\text{ }=\text{ }ab\text{ }sin\text{ }C\]

\[2bc\text{ }sin\text{ }A\text{ }=\text{ }2ac\text{ }sin\text{ }B\text{ }=\text{ }2ab\text{ }sin\text{ }C\]

∴ From equation (i), (ii) and (iii) we get,

\[({{c}^{2}}~\text{ }{{a}^{2}}~+\text{ }{{b}^{2}})\text{ }tan\text{ }A\]

\[=\text{ }({{a}^{2}}~\text{ }{{b}^{2}}~+\text{ }{{c}^{2}})\text{ }tan\text{ }B\text{ }=\text{ }({{b}^{2}}~\text{ }{{c}^{2}}~+\text{ }{{a}^{2}})\text{ }tan\text{ }C\]

Hence proved.

i

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