Match each entry in List - I to the correct entries in List - II LIST-I LIST-III)The value of (1+tan8°)(1+tan37°)(1+tan22°)(1+tan23°)P)6II)In a triangle ABC ∑cotA+cotBtanA+tanB+2 Q)4III)In a triangle 3sinA+4cosB=6,3cosA+4sinB=1 then ∠C=K×5° . The value of KR)3IV)The value of log10tan1°+log10tan2°+.....+log10tan89°+4S)1

I. A+B=45° (1+tanA)(1+tanB)=2II. ∑cotA+cotBtanA+tanB A+B+C=180° cotAcotB+cotBcotC+cotC+cotA=1 1+2=3III. Squaring and Adding 9+16+24sin(A+B)=37 24sin(A+B)=12sin(A+B)=12 ⇒A+B is 30° or 150° ⇒C=30°∠C=K×5°K=6 IV. log10tan1°+log10tan2°+...+log10tan89° A+B=90°tanA=cotB log10tan1°+log10tan2°+...+logtan1089°+4 0+4=4

Match each entry in List - I to the correct entries in List - II LIST-I LIST-III)The value of (1+tan8°)(1+tan37°)(1+tan22°)(1+tan23°)P)6II)In a triangle ABC ∑cotA+cotBtanA+tanB+2 Q)4III)In a triangle 3sinA+4cosB=6,3cosA+4sinB=1 then ∠C=K×5° . The value of KR)3IV)The value of log10tan1°+log10tan2°+.....+log10tan89°+4S)1

I. A+B=45° (1+tanA)(1+tanB)=2II. ∑cotA+cotBtanA+tanB A+B+C=180° cotAcotB+cotBcotC+cotC+cotA=1 1+2=3III. Squaring and Adding 9+16+24sin(A+B)=37 24sin(A+B)=12sin(A+B)=12 ⇒A+B is 30° or 150° ⇒C=30°∠C=K×5°K=6 IV. log10tan1°+log10tan2°+...+log10tan89° A+B=90°tanA=cotB log10tan1°+log10tan2°+...+logtan1089°+4 0+4=4

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Match each entry in List - I to the correct entries in List - II
LIST-I LIST-II
I) The value of $\frac{(1 + \tan 8 °) (1 + \tan 37 °)}{(1 + \tan 22 °) (1 + \tan 23 °)}$ P) 6
II) In a triangle $A B C$ $\sum \frac{\cot A + \cot B}{\tan A + \tan B} + 2$ Q) 4
III)
In a triangle $3 \sin A + 4 \cos B = 6, 3 \cos A + 4 \sin B = 1$ then $∠ C = K \times 5 °$ .
The value of $K$
R) 3
IV) The value of $\log_{10} \tan 1 ° + \log_{10} \tan 2 ° + ..... + \log_{10} \tan 89 ° + 4$ S) 1

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answer is A.

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Detailed Solution

I. $A + B = 45 °$
$(1 + \tan A) (1 + \tan B) = 2$
II. $\sum \frac{\cot A + \cot B}{\tan A + \tan B}$
$A + B + C = 180 ° \cot A \cot B + \cot B \cot C + \cot C + \cot A = 1 1 + 2 = 3$
III. Squaring and Adding
$9 + 16 + 24 \sin (A + B) = 37$
$\begin{matrix} 24 \sin (A + B) = 12 & \sin (A + B) = \frac{1}{2} \end{matrix} \Rightarrow A + B is 30 ° or 15 0 ° \Rightarrow \begin{matrix} C = 30 ° & ∠ C = K \times 5 ° & K = 6 \end{matrix}$
IV. $\log_{10} \tan 1 ° + \log_{10} \tan 2 ° + ... + \log_{10} \tan 89 °$
$\begin{matrix} A + B = 90 ° & \tan A = \cot B \end{matrix} \log_{10} \tan 1 ° + \log_{10} \tan 2 ° + ... + {logtan}_{10} 89 ° + 4 0 + 4 = 4$

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