Question

pls answer this question correctly
Chapter=Trigonometry​

Answer 1

Question:-

To prove the identity.

Identity:-

$\sf{\dfrac{1}{sec\theta - tan\theta} + \dfrac{1}{sec\theta + tan\theta} = \dfrac{2}{cos\theta}}$

Solution:-

Taking LHS,

$\sf{\dfrac{1}{sec\theta - tan\theta} + \dfrac{1}{sec\theta + tan\theta}}$

= $\sf{\dfrac{1}{\dfrac{1}{cos\theta} - \dfrac{sin\theta}{cos\theta}} + \dfrac{1}{\dfrac{1}{cos\theta} + \dfrac{sin\theta}{cos\theta}}}$

= $\sf{\dfrac{1}{\dfrac{1-sin\theta}{cos\theta}} + \dfrac{1}{\dfrac{1+sin\theta}{cos\theta}}}$

= $\sf{\dfrac{cos\theta}{1-sin\theta} + \dfrac{cos\theta}{1+sin\theta}}$

Taking LCM

$\sf{\dfrac{cos\theta(1+sin\theta) + cos(1-sin\theta)}{(1-sin\theta)(1+sin\theta)}}$

= $\sf{\dfrac{cos\theta + cos\theta sin\theta + cos\theta - cos\theta sin\theta}{1-sin^2\theta}}$

= $\sf{\dfrac{2cos\theta + \cancel{cos\theta sin\theta} - \cancel{cos\theta sin\theta}}{cos^2\theta}}$

= $\sf{\dfrac{2cos\theta}{cos^2\theta}}$

= $\sf{\dfrac{2}{cos\theta} = [RHS]}$

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Extra Information!!

Points to be remembered:-

• $\sf{sin\theta = \dfrac{1}{cosec\theta}}$

• $\sf{cosec\theta = \dfrac{1}{sin\theta}}$

• $\sf{cos\theta = \dfrac{1}{sec\theta}}$

• $\sf{sec\theta =\dfrac{1}{cos\theta}}$

• $\sf{tan\theta = \dfrac{1}{cot\theta}}$

• $\sf{cot\theta = \dfrac{1}{tan\theta}}$
• Also, $\sf{tan\theta = \dfrac{sin\theta}{cos\theta}}$

• $\sf{cot\theta = \dfrac{cos\theta}{sin\theta}}$

Some other important identities:-

• $\sf{sin^2\theta + cos^2\theta = 1}$
• $\sf{\implies sin^2\theta = 1-cos^2\theta}$
• $\sf{\implies cos^2\theta = 1-sin^2\theta}$
• $\sf{1+tan^2\theta = sec^2\theta}$
• $\sf{\implies sec^2\theta - tan^2\theta = 1}$
• $\sf{\implies sec^2\theta - 1 = tan^2\theta}$
• $\sf{1+cot^2\theta = cosec^2\theta}$
• $\sf{\implies cosec^2\theta - cot^2\theta = 1}$
• $\sf{\implies cosec^2\theta - 1 = cot^2\theta}$

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