Hola Brainlians. Here’s a question to check your Trigonometric skill. ICSE Board : Trigonometry, Class 10. Give answer fast and n

2 thoughts on “Hola Brainlians. Here’s a question to check your Trigonometric skill. ICSE Board : Trigonometry, Class 10. Give answer fast and n”

1. Step-by-step explanation:

   We will use simplification in both the sides, for easy process underway.

   So, as hinted, simplifying the RHS will help.

   We have, RHS as :-

   [tex]\sf{\dfrac{1}{tanx+cotx}}[/tex]

   We can also write tanx and cotx in terms of sinx and cosx.

   [tex]\sf{\dfrac{1}{\dfrac{sinx}{cosx}+ \dfrac{cosx}{sinx}}}[/tex]

   [tex]\sf{\dfrac{1}{\dfrac{sin^2 x + cos^2 x}{cosx . sinx}}}[/tex]

   sin²x + cos²x = 1 , applying it:-

   [tex]\sf{\dfrac{1}{\dfrac{1}{cosx . sinx}}}[/tex]

   Resultant RHS – [tex]\sf{cosx . sinx }[/tex]

   Moving on to LHS,

   [tex]\sf{(\dfrac{1}{sinx} – sinx)(\dfrac{1}{cosx} – cosx)}[/tex]

   Converted cosecx into 1/sinx and secx into 1/cosx

   [tex]\sf{(\dfrac{1 – sin^2 x}{sinx})(\dfrac{1 – cos^2 x}{cosx})}[/tex]

   [tex]\sf{\dfrac{(1 – sin^2 x)(1 – cos^2 x)}{sinx . cosx}}[/tex]

   Breaking the brackets,

   [tex]\sf{\dfrac{1 – cos^2 x – sin^2 x + sin^2x . cos^2 x }{sinx . cosx}}[/tex]

   Applying sin²x + cos²x = 1 ,

   [tex]\sf{\dfrac{1 – 1 sin^2x . cos^2 x }{sinx . cosx}}[/tex]

   [tex]\sf{\dfrac{(sinx . cosx)(sinx . cosx) }{sinx . cosx}}[/tex]

   Cutting one of the brackets off with denominator,

   We get cosx . sinx

   cosx . sinx = cosx . sinx

   LHS = RHS

   Hence, proved.

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2. Question:–

   • Proof the given identity.[tex]\sf (cosecx – sinx)(secx – cosx) = \dfrac { 1 } { (tanx + cotx) } [/tex]

   To Find:–

   • Prove that.

   Solution:–

   We have RHS as:-

   • [tex]\sf \: \dfrac { 1 } { tanx + cotx } [/tex]

   Now ,

   [tex]\longrightarrow\sf \: \dfrac { 1 } { \dfrac { sinx } { cosx } + \dfrac { cosx } { sinx } } [/tex]

   Here ,

   • [tex]\sf \: { sin }^{ 2 } x + { cos }^{ 2 } x = 1 [/tex]

   [tex]\longrightarrow\sf \: \dfrac { 1 } { \dfrac { 1 } { cosx.sinx } } [/tex]

   Now ,

   [tex]\longrightarrow\sf \: LHS = ( \dfrac { 1 } { sinx } – sinx )( \dfrac { 1 } { cosx } – cosx ) [/tex]

   [tex]\longrightarrow\sf \: \dfrac { 1 – { cos }^{ 2 }x – { sin }^{ 2 }x + { sin }^{ 2 }x \times { cos }^{ 2 }x }{ sinx \times cosx } [/tex]

   [tex]\longrightarrow\sf \: \dfrac { ( sinx.cosx )( sinx.cosx ) } { ( sinx.cosx ) } [/tex]

   [tex]\longrightarrow\sf \: sinx.cosx [/tex]

   [tex]\longrightarrow\sf \: LHS = RHS [/tex]

   • Hence proved

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