RRB NTPC Mains or Stage 2 Model Paper 6 with solutions

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Question : 18 of 119

Marks: +1, -0

If cos A = tan B, cos B = tan C and cos C = tan A then

sin^2

A is equal to

${√ 5 –1}/ 4$
${√ 5 –1} /2$
${3-√ 5} /2$
${√3 –1} /a42$

Solution:

(c) CosA = tanB
Squaring on both sides

cos^2A = tan^2B

tan^2B= {sin^2B}/{cos^2B} ={1-cos^2B}/ {cos^2 B}

∴ cos^2 A = {1- cos^2B}/ {cos^2B}

cos^2A={1- tan^2C} /{tan^2C}

(∵ cosB=tanC)

⇒cos^2Atan^2C=1-tan^2C=1-{sin^2C}/{cos^2C}

⇒cos^2A(1-cos^2C)=2cos^2C-1

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