Saturday, 29 June 2019

CBSE Class 10 - Maths - Trigonometry Question - If a secπœƒ + b tanπœƒ + c = 0 and p secπœƒ + q tanπœƒ + r = 0, prove that: (π‘π‘Ÿ − π‘žπ‘)² − (𝑝𝑐 − π‘Žπ‘Ÿ)² = (π‘Žπ‘ž − 𝑏𝑝)²(#eduvictors)

CBSE Class 10 Maths: Trigonometry Question

CBSE Class 10 - Maths - Trigonometry Question - If a secπœƒ + b tanπœƒ + c = 0 and      p secπœƒ + q tanπœƒ + r = 0,  prove that:  (π‘π‘Ÿ − π‘žπ‘)² − (𝑝𝑐 − π‘Žπ‘Ÿ)² = (π‘Žπ‘ž − 𝑏𝑝)²(#eduvictors)

Question: If a secπœƒ + b tanπœƒ + c = 0 and  p secπœƒ + q tanπœƒ + r = 0,  prove that:  
                                  (π‘π‘Ÿ − π‘žπ‘)² − (𝑝𝑐 − π‘Žπ‘Ÿ)² = (π‘Žπ‘ž − 𝑏𝑝)²

Answer:

a secπœƒ + b tanπœƒ + c = 0 
c = - (a secπœƒ + b tanπœƒ)

and p secπœƒ + q tanπœƒ + r = 0
⇒  r = -(p secπœƒ + q tanπœƒ)

LHS = (π‘π‘Ÿ − π‘žπ‘)² − (𝑝𝑐 − π‘Žπ‘Ÿ)²

= [b(-p secπœƒ - q tanπœƒ) - q(-a secπœƒ - b tanπœƒ)]² - [p(-a secπœƒ - b tanπœƒ) - a(-p secπœƒ - q tanπœƒ)]²

= [-bp secπœƒ - bq tanπœƒ + aq secπœƒ + bq tanπœƒ ]² - [-ap secπœƒ - bp tanπœƒ + ap secπœƒ + aq tanπœƒ]²

= [aq secπœƒ - bp secπœƒ]² - [aq tanπœƒ - bp tanπœƒ]²

= (aq - bp)²sec²πœƒ - (aq - bp)²tan²πœƒ

= (aq - bp)²[sec²πœƒ - tan²πœƒ]

= (aq - bp)²                 [∵ sec²πœƒ - tan²πœƒ = 1]

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