Question

*Trigonometry Class 10*

RD SHARMA (12th Revised 2019)

Exe.11.1

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Q.68. Prove that:

(cosecA - secA)(cotA - tanA) = (cosecA + secA)(secA.cosecA - 2)

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Q.74 Prove that:


$\dfrac{cot^2A(secA-1)}{1 + sinA} = sec^2A \left( \dfrac{1 -sinA}{1 + secA} \right)$


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Thanks in advance!

Answer 1

Hi there !!

Look at the attachment ↑↑

Identity used :-

a²-b²=(a+b)(a-b)

(a-b)² = a²+b²-2ab

sin²0+ cos²0=1

cos²0 = 1- sin²0

cosec0 = 1/sin0

sec0= 1/cos0

cot0 = cos0/sin0

tan0 = sin0/cos0

Thankyou :)

Answer 2

Q 68. (cosecA - secA)(cotA - tanA) = (cosecA + secA)(secA cosecA - 2)

Proof: We will solve the R.H.S first.

R.H.S → (cosecA + secA)(secA cosecA - 2)

R.H.S → (1/sinA + 1/cosA)(1/cosA · 1/sinA - 2)

R.H.S → (cosA + sinA)/sinA cosA × (1 - 2 sinA cosA)/sinA cosA

R.H.S → (cosA + sinA)/sinA cosA × (cosA - sinA)²/sinA cosA

R.H.S → (cosA + sinA)(cosA - sinA)(cosA - sinA)/sin²A cos²A

Now solving L.H.S we get,

L.H.S → (cosecA - secA)(cotA - tanA)

L.H.S → (1/sinA - 1/cosA)(cosA/sinA - sinA/cosA)

L.H.S → (cosA - sinA)/sinA cosA × (cos²A - sin²A)/sinA cosA

L.H.S → (cosA - sinA)(cosA + sinA)(cosA - sinA)/sin²A cos²A

Since L.H.S = R.H.S

Hence Proved

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Q 74. cot²A(secA - 1)/(1 + sinA) = sec²A(1 - sinA)/(1 + secA)

Proof: Given in attachment

Remember that:

• Identities has been used
• (1 - cos²A) = (1 + cosA)(1 - cosA)
• (1 - sin²A) = (1 + sinA)(1 - sinA)